I came up with the following proof of the above theorem which is the key to the Galois's theory of algebraic equations. The usual proof uses Lagrange resolvent or Hilbert 90 which uses a similar trick. This proof is conceptual. Is this proof well-known?
Proposition
Let $K$ be a field. Let $n$ be a positive integer which is not divisible by the characteristic of $K$. Suppose $K$ contains an $n$-th primitive root of unity. Let $L$ be a cyclic extension of degree n over a field $K$. Then there is an element $y$ of $L$ such that $L = K(y)$ and $y^n$ is an element of $K$.
Proof: Let $\sigma$ be a generator of the Galois group of $L/K$. It is well-known that $1,\sigma,\dotsc,\sigma^{n-1}$ are linearly independent over $K$. Hence, since $\sigma^n = 1$, $X^n - 1$ is the minimal polynomial of $\sigma$ over $K$. Let $f(x)$ be the characteristic polynomial of $\sigma$. By the Cayley-Hamilton theorem, $f(\sigma) = 0$. Hence $f(X)$ is divisible by $X^n - 1$. Since $f(X)$ is monic and the degree of $f(X)$ is $n$, $f(X) = X^n - 1$. Let $\zeta$ be an $n$-th primitive root of unity. Since $f(\zeta) = 0$, $\zeta$ is an eigenvalue of $\sigma$. Hence there is an element $y \neq 0$ of $L$ such that $\sigma(y) = \zeta y$. Since $\sigma(y^n) = (\zeta y)^n = y^n$, $y^n$ is an element of $K$ by the fundamental theorem of Galois theory. $\sigma(y^i) = (\zeta y)^i = (\zeta^i)y^i,i = 0,1,\dotsc,n - 1$. Hence $1,y,\dotsc,y^{n-1}$ are eigenvectors of $1,\zeta,\dotsc,\zeta^{n-1}$ respectively. Hence $1,y,\dotsc,y^{n-1}$ are linearly independent over K. Hence $L = K(y)$.