Let $K$ be a field of characteristic $0$.
Let $p$ be a prime number. Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$ and $X^p - \alpha^p$ is irreducible over $K$. Then we call $K(\alpha)/K$ a simple prime radical extension.
Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields. If $K_i/K_{i-1}$ is a simple prime radical extension for $i =1, \dots, n$, we call $L/K$ is a prime radical extension.
Let $L/K$ be a field extension. If there exists a prime radical extension $E/K$ such that $L \subset E$, we call $L/K$ a prime radically solvable extension.
Lemma 1 Let $K$ be a field. Let $p$ be a prime number such that $char(K) \neq p$. Let $a \in K$. Suppose $X^p - a$ has no roots in $K$. Then $X^p - a$ is irreducible over $K$.
Proof: Let $\bar K$ be an algebraic closure of $K$. Let $\zeta$ be a $p$-th root of unity in $\bar K$. Let $\alpha$ be a root of $X^p - a$ in $\bar K$. Then $\alpha, \alpha\zeta, \dots, \alpha\zeta^{p-1}$ are distinct roots of $X^p - a$. Suppose $X^p - a$ is not irreducible over $K$. Then $X^p - a = g(X)h(X)$, where $g(X)$ and $h(x)$ are monic irreducible polynomials of degree $> 0$. Suppose deg $g(X) = k$ and $b$ is the constant term of $g(X)$. Then $b = \pm\alpha^k\zeta^m$ for some integer $m$. Hence $b^p = \pm a^k$ if $p$ is odd, and $b^p = a^k$ if $p = 2$. Hence there exists $c \in K$ such that $c^p = a^k$. Let $K^*$ be the multiplicative group of $K$. Let $(K^*)^p = \{x^p| x \in K^*\}$. Let $\gamma$ be the image of $a$ by the canonical homomorphism $K^* \rightarrow K^*/(K^*)^p$. Then $\gamma \neq 1$ and $\gamma^p = 1$. Hence the order of $\gamma$ is $p$. Since $c^p = a^k$, $\gamma^k = 1$. This is a contradiction. QED
Lemma 2 Let $K$ be a field. Let $p$ be a prime number such that $char(K) \neq p$. Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$. Let $\alpha^p = a$. Then one of the following three cases occurs.
1) $X^p - a$ is irreducible over $K$.
2) $K(\alpha) = K(\zeta)$, where $\zeta$ is a primitive $p$-th root of unity.
3) $K(\alpha) = K$.
Proof: Let $\bar K$ be an algebraic closure containing $\alpha$. Let $\zeta$ be a primitive $p$-th root of unity in $\bar K$. Suppose $X^p - a$ is not irreducible. By Lemma 1, there exists $b \in K$ such that $b^p = a$. Since $b, b\zeta,\dots,b\zeta^{p-1}$ are all the roots of $X^p - a$, $\alpha = b\zeta^i, 0 \le i \le p -1$. If $i = 0, K(\alpha) = K$. If $i > 0$, $\zeta^i$ is a primitive $p$-th root of unity. Hence $K(\alpha) = K(b\zeta^i) = K(\zeta^i) = K(\zeta)$. QED
Lemma 3 Let $K$ be a field. Let $n > 0$ be an integer which is not divisible by $char(K)$. Suppose $K$ contains a primitive $n$-th root $\zeta$ of unity. Let $a \in K$. Let $\alpha$ be a root of $X^n - a$ in an algebraic closure of $K$. Then $K(\alpha)/K$ is a cyclic extension and $(K(\alpha)\colon K)$ is a divisor of $n$.
Proof: If $a = 0$, the assertion is trivial. Hence we assume $a \neq 0$. Hence $\alpha \neq 0$. Hence $\alpha, \alpha\zeta,\dots,\alpha\zeta^{n-1}$ are the distincts roots of $X^n - a$. Since $\zeta \in K, K(\alpha, \alpha\zeta,\dots,\alpha\zeta^{n-1}) = K(\alpha)$. Hence $K(\alpha)/K$ is a Galois extension. Let $G$ be the Galois group of $K(\alpha)/K$. Let $\Gamma$ be the set of $n$-th roots of unity in $K$. $\Gamma$ is a cyclic group of order $n$. Let $\sigma \in G$. Since $\sigma(\alpha)$ is a root of $X^n - a$, there exists a unique $\omega \in \Gamma$ such that $\sigma(\alpha) = \alpha\omega$. Hence we get a map $\psi\colon G \rightarrow \Gamma$ such that $\sigma(\alpha) = \alpha\psi(\alpha)$.It is easy to see that $\psi$ is an injective homomorphism. QED
Lemma 4 Let $\Omega$ be an algebraically closed field of characteristic $0$. Let $p$ be a prime number. Let $L/K$ be a prime radical extension of degree $p$, where $L$ is a subfield of $\Omega$. Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$. Let $F$ be a subfield of $\Omega$ containing $\zeta$. Then $FL/FK$ is a trivial extension or a simple prime radical extension of degree $p$.
Proof: Since $L/K$ is a prime radical extension of degree $p$, there exists $\alpha \in L$ such that $L = K(\alpha)$ and $\alpha^p \in K$. Then $FL = FK(\alpha)$ and $\alpha^p \in K \subset FK$. By Lemma 3, $FL/FK$ is a trivial extension or a cyclic extension of degree $p$. Suppose $FL/FK$ is a cyclic extension of degree $p$. Then $X^p - a$ is the minimal polynomial of $\alpha$ over $FK$, where $a = \alpha^p$. Hence $X^p - a$ is irreducible over $FK$. Hence $FL/FK$ is a simple prime radical extension of degree $p$. QED
Lemma 5 Let $\Omega$ be an algebraically closed field of characteristic $0$. Let $L/K$ be a prime radical extension of degree $n$, where $L$ is a subfield of $\Omega$. Let $p_1,\dots,p_r$ be all the distinct prime factors of $n$. Let $m = \prod_i p_i$. Let $\zeta$ be a primitive $m$-th root of unity in $\Omega$. Let $F$ be a subfield of $\Omega$ containing $\zeta$. Then $FL/FK$ is a prime radical extension and $(FL \colon FK)$ is a divisor of $n$.
Proof: Since $L/K$ is a prime radical extension, there exists a tower of subfields of L: $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ such that each $K_i/K_{i-1}$ is a simple radical extension. Since $n = \prod_i (K_i\colon K_{i-1})$, each $(K_i\colon K_{i-1})$ is one of the primes $p_1, \dots,p_r$. Hence $F$ contains a primitive $p_i$-th unity for each $i$. By Lemma 4, $FK_i/FK_{i-1}$ is trivial or a simple prime radical extension. Hence the assertions follow. QED
Lemma 5.5 Let $\Omega$ be an algebraically closed field. Let p be a prime number such that $p \neq char(\Omega)$. Let $L/K$ be a cyclic extension of degree $p$, where $L$ is a subfield of $\Omega$. Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$. Then $L(\zeta)/K(\zeta)$ is a trivial or a simple prime radical extension.
Proof: Since $L(\zeta) = LK(\zeta)$, $Gal(L(\zeta)/K(\zeta)$) is isomorphic to a subgroup of $Gal(L/K)$. Hence $|Gal(L(\zeta)/K(\zeta))| = 1$ or $p$. Suppose $|Gal(L(\zeta)/K(\zeta))| = p$. Then by this theorem, $L(\zeta)/K(\zeta)$ is a simple prime radical extension. QED
Lemma 6 Let $K$ be a field of characteristic $0$. Let $\zeta$ be a primitive $n$-th root of unity in an algebraic closure $\bar K$ of $K$. Then $K(\zeta)/K$ is a prime radically solvable extension.
Proof: We use induction on $n$. Since the assertion of the lemma is clear if $n = 1$, we assume that $n > 1$. Let $G$ be the Galois group of $L/K$, where $L = K(\zeta)$. $G$ is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z})^\times$. Hence, by Galois theory, there exists a tower: $K = K_0 \subset K_1 \subset \cdots \subset K_r = L$ such that each $K_i/K_{i-1}$ is a cyclic extension of a prime degree. Let $p_1,\dots,p_s$ be all the distinct prime divisors of $|G|$. Since $|G| = \prod_i (K_i \colon K_{i-1})$, each $(K_i \colon K_{i-1})$ is one of $p_1,\dots,p_s$. Let $m = \prod_j p_j$. Let $\eta$ be a $m$-th primitive root of unity in $\bar K$. By Lemma 5.5, $K_i(\eta)/K_{i-1}(\eta)$ is trivial or a simple prime radical extension. Hence, $L(\eta)/K(\eta)$ is a prime radical extension. Since $m \le |G| \le |(\mathbb{Z}/n\mathbb{Z})^*| < n$, by the induction hypothesis, there exists a prime radical extension $E/K$ such that $K(\eta) \subset E$. By Lemma 5, $EL(\eta)/EK(\eta) = EL/EK = EL/E$ is a prime radical extension. Hence $EL/K$ is a prime radical extension. Since $L \subset EL$, $L/K$ is a prime radically solvable extension. QED
Lemma 7 Let $\Omega$ be a field of characteristic $0$. Let $K$ be a subfield of $\Omega$. Let $L/K$ be a simple prime radical subextension of $\Omega/K$. Let $F/K$ be a subextension of $\Omega/K$. Then $FL/F$ is a prime radically solvable extension.
Proof: Since $L/K$ is a simple prime radical extension, there exist a prime number $p$ and $\alpha \in L$ such that $L = K(\alpha)$ and $\alpha^p \in K$. Then $FL = F(\alpha)$. By Lemma 2 and Lemma 6, the assertion follows. QED
Lemma 8 Let $\Omega$ be an algebraically closed field of characteristic $0$. Let $K \subset M \subset L$ be a tower of subfields of $\Omega$. Suppose $M/K$ is a prime radically solvable extension and $L/M$ is a simple prime radical extension. Then $L/K$ is a prime radically solvable extension.
Proof: There exists a tower $K \subset M \subset F$ of subfields of $\Omega$ such that $F/K$ is a prime radical extension. By Lemma 7, $FL/F$ is a prime radically solvable. Hence $FL/K$ is also a prime radically solvable extension. Hence the assertion follows. QED
Lemma 9 Let $K \subset M \subset L$ be a tower of fields. Suppose $M/K$ is a prime radically solvable extension and $L/M$ is a prime radical extension. Then $L/K$ is a prime radically solvable extension.
Proof: This follows immediately from Lemma 8.
Lemma 10 Let $K \subset M \subset L$ be a tower of fields. Suppose $M/K$ and $L/M$ are prime radically solvable extensions. Then $L/K$ is also a prime radically solvable extension.
Proof: This follows immediately from Lemma 9.
Lemma 11 Let $K$ be a field of characteristic $0$. Let $n > 1$ be an integer. Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^n \in K$. Then $K(\alpha)/K$ is a prime radically solvable extension.
Proof: We use induction on $n$. If $n = 2$, the assertion is clear. We assume $n > 2$. Let $p$ be a prime divisor of $n$. Let $m = \frac{n}{p}$. Since $(\alpha^p)^m = \alpha^n \in K$, by the induction hypothesis, $K(\alpha^p)/K$ is a prime radically solvable extension. By Lemma 2 and Lemma 6, $K(\alpha)/K(\alpha^p)$ is a prime radically solvable extension. Hence, by Lemma 10, $K(\alpha)/K$ is a prime radically solvable extension. QED
Let $K$ be a field of characteristic $0$. Let $L/K$ be an extension. Let $n > 0$ be an integer. Suppose there exists an element $\alpha$ of $L$ such that $L = K(\alpha)$ and $\alpha^n \in K$. Then we call $L/K$ is a simple radical extention.
Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields. If $K_i/K_{i-1}$ is a simple radical extension for for $i =1, \dots, n$, we call $L/K$ is a radical extension.
Let $M/K$ be an extention. If there esists a radical extension $L/K$ such that $M$ is a subfield of $L$, $M/K$ is called a radically sovable extension.
Lemma 12 Let $\Omega/K$ be an extension of a field $K$ of characteristic $0$. Let $L/K$ be a radically solvable subextension of $\Omega/K$. Let $F/K$ be a subextension of $\Omega/K$. Then $FL/F$ is radically solvable.
Proof: There exists a tower: $K = K_0 \subset K_1 \subset \cdots \subset K_n = E$ such that $L \subset E$, where $K_i/K_{i-1}$ is a simple radical extension for for $i =1, \dots, n$ Then $F = FK_0 \subset FK_1 \subset \cdots \subset FK_n = FE$. Since $FK_i/FK_{i-1}$ is a simple radical extension, $FL/F$ is radically solvable. QED
Lemma 13 Let $K$ be a field of characteristic $0$. Let $K \subset M \subset L$ be a tower of fields. Suppose $M/K$ and $L/M$ are radically solvable. Then $L/K$ is also radically solvable.
Proof: There exists a tower: $K = K_0 \subset K_1 \subset \cdots \subset K_n = F$ such that $M \subset F$, where $K_i/K_{i-1}$ is a simple radical extension for for $i =1, \dots, n$ By Lemma 12, $FL/F$ is radically solvable. Hence $L/K$ is radically solvable. QED
Lemma 14 Let $K$ be a field of characteristic $0$. Let $x \in K^{rad}$. Then $K(x)/K$ is radically solvable.
Proof: For each integer $i \ge 1$, let $K_i$ be the subfield of $\bar K$ as defined by the title question. Let $r$ be the least integer such that $x \in K_r$. We use induction on $r$. If $r = 1$, the assertion is clear. Suppose $r > 1$. Then there exists a positive integer $n$ such that $x^n \in K_{r-1}$. By the induction hypothesis, $K(x^n)/K$ is radically solvable. Clearly $K(x)/K(x^n)$ is radically solvable. Hence, by Lemma 13, $K(x)/K$ is radically solvable. QED
Lemma 15 A radically solvable extension $L/K$ is a prime radically solvable extension.
Proof: This follows immediately from Lemma 10 and Lemma 11.
Proposition Let $K$ be a field of characteristic $0$. Let $x \in K^{rad}$. Then $K(x)/K$ is prime radically solvable.
Proof: This follows immediately from Lemma 14 and Lemma 15.