I've solved the following problem but I have some doubts about my proof.
Problem statement
$K$ is a field. Given $p(x)\in K[x]$ an irreducible and separable polynomial over $K$ and $E$ its splitting field over $K$, prove that $E/K$ is Galois.
Proof
We denote $n=deg(p)$. As $p(x)=\lambda\,(x-\alpha_1)\cdots(x-\alpha_n)$ with $\lambda\in K$ and $\alpha_i$ the roots of $p(x)$ in $E$, by the minimal property of the splitting field we have that $E=K\left(\alpha_1,\ldots,\alpha_n\right)$.
We write $K_0=K$ and $K_i = K_{i-1}(\alpha_i)$ for $i=1,\ldots,n$ and we consider the following chain of fields:
$K=K_0 \subset K_1 \subset \ldots \subset K_{n-1} \subset K_n = E.$
So we will prove the statement by induction over $i$ by showing that $[K_i:K]$ is the number of morphisms $\varphi_i:\,K_i\longrightarrow E$ that $\varphi_i\mid_K=id$ for $i=1,\ldots,n$.
If $i=1$, then $p(x)=Irr(\alpha_1,K)$ and the number of morphisms $\varphi_1$ is defined by the number of roots of $\underline{\varphi_1}\left(Irr(\alpha_1,K)\right)=\underline{id}(p(x))=p(x)$ (where $\underline{\varphi_i}:K_i[x]\longrightarrow E[x]$ sends a polynomial $q(x) = a_0+a_1\,x+\ldots+a_m\,x^m$ to $\underline{\varphi_i}\left(q(x)\right):=\varphi_i(a_0)+\varphi_i(a_1)\,x+\ldots+\varphi_i(a_m)\,x^m$).
Since $p(x)$ is irreducible and separable all roots are distinct and the number of roots is
$n=deg(p)=deg(Irr(\alpha_1,\,K)) = [K(\alpha_1):K] = [K_1:K].$
Now we assume that it is satisfied for $i=n-1$ and we prove it for $i=n$.
As $Irr(\alpha_n,K_{n-1})\mid Irr(a_n,K)$, we have that
$\underline{\varphi_{n-1}}\left(Irr(\alpha_n,K_{n-1})\right)\mid\underline{\varphi_{n-1}}\left(Irr(a_n,K)\right)= \underline{id}\left(p(x)\right) = p(x).$ Since $p(x)$ splits over $E$, $\underline{\varphi_{n-1}}\left(Irr(\alpha_n,K_{n-1})\right)$ also splits over $E$ and the number of roots in $E$ is
$deg\left(\underline{\varphi_{n-1}}\left(Irr(\alpha_n, K_{n-1})\right)\right) = deg(Irr(\alpha_n, K_{n-1})) = [K_n:K_{n-1}].$
Thus, the number of ways of defining $\varphi_n:\,K_n\longrightarrow E$ that satisfies $\varphi_n\mid_{K_{n-1}} = \varphi_{n-1}$ is $[K_n:K_{n-1}]$.
Finally, applying the induction hypothesis and the formula for degrees, we have
$\left|\right\{\varphi_n:K_n\rightarrow E\mid \varphi_n\mid_K =id\left\}\right| = [K_n:K_{n-1}]\,[K_{n-1}:K] = [K_n:K] = [E:K].$
Doubts
First of all, it is correct to assume $E=K\left(\alpha_1,\ldots,\alpha_n\right)$? I've found some texts that refer to various splitting fields for the same polynomial. However, I thought that there was a unique splitting field, the one that satisfies the minimal property. Where is my mistake?
Also, there is a step in the proof that I borrowed from another proof. It's when it says that since $\underline{\varphi_{n-1}}\left(Irr(\alpha_n,K_{n-1})\right)$ divides $p(x)$ it also splits in $E$. I don't really understand this step.
Finally, do you see any other mistake?