Splitting Fields. Prove Q(alpha^4)=Q(alpha)

[caffeinemachine]

Let $f(x)\in \mathbb Q[x]$ be an irreducible polynomial of degree $n\geq 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Given that $[L:\mathbb Q]=n!$, prove that $\mathbb Q(\alpha^4)=\mathbb Q(\alpha)$.
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Attempt:

Lemma:
Let $F$ be any field and $f,g \in F[x]$. Let $K$ and $L$ be splitting fields for $f$ and $g$ over $F$ respectively. Write $h=fg\in F[x]$. Let $E$ be the splitting field for $h$ over $F$. Then $[E:F]\leq [K:F][L:F]$

Proof: Let $f=(x-\alpha_1)\cdots(x-\alpha_k)$ in $K[x]$ and $g=(x-\beta_1)\cdots(x-\beta_l)$ in $L[x]$. Its easy to see that $K=F(\alpha_1,\ldots,\alpha_k)$, $L=F(\beta_1,\ldots,\beta_l)$ and $E=F(\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_l)$. From here its easy to conclude that \begin{equation*}[E:K]\leq [L:F]\tag{1}\end{equation*}Now, $[E:F]=[E:K][K:F]\leq [K:F][L:F]$. Hence the lemma is settled.$\blacksquare$

Corollary: Let $F$ be any field and $f\in F[x]$ with deg $f=n$. Let $E$ be the spliting field for $f$ over $F$. Assume that $[E:F]=n!$. Then $f$ is irreducible over $F$.

Assuming there are no mistakes in my argument..
In the light of the corollary I think the question redundantly confers irreducibility to $f$.

Now. Let $f=(x-\alpha_1)\cdots (x-\alpha_n)=(x-\alpha_1)g$ be the representation of $f$ in $L[x]$. Define $F_1=\mathbb Q(\alpha_1)$. Then $L$ is the splitting field for $g$ over $F_1$ satisfying $[L:F_1]=(n-1)!=\text{ deg }g$ and hence $g$ is irreducible over $F_1$. This piece of information suggests that induction might be helpful. I can settle the base case, that is, $n=3$. If you require then I can post the proof for the base case in my following posts.
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Can anybody see how to settle the question?

 

[jvicens]

Let $[L: \mathbb{Q}] = n!$ and let $K = \mathbb{Q}(\alpha)$. Then from the given information, we have $[K:\mathbb{Q}] = n$. Since $\alpha$ is a zero of the irreducible polynomial $f$, we know that $K$ is a splitting field for $f$ over $\mathbb{Q}$. By the fundamental theorem of Galois theory, we have $[L:K] = [L:\mathbb{Q}] / [K:\mathbb{Q}] = (n!)/n = (n-1)!$. This means that $L$ is a splitting field for $f$ over $K$ with degree $(n-1)!$. Since $f$ is irreducible over $K$, we must have $f = (x-\alpha)g$ for some irreducible polynomial $g \in K[x]$ with degree $n-1$. This implies that $g$ is the minimal polynomial for $\alpha$ over $K$.
Now, let $M = K(\alpha^4)$. We want to show that $M = K(\alpha)$. Since $\alpha \in L$, we have $M \subseteq L$. Also, since $\alpha^4 \in K$, we have $K \subseteq M$. Therefore, we have $K \subseteq M \subseteq L$. We want to show that $[M:K] = 1$, which will imply that $M = K(\alpha)$.
Since $M$ is a finite extension of $K$, we have $[M:K] = [M:L][L:K]$. Since $L$ is a splitting field for $g$ over $K$, we know that $g$ splits completely in $L$. This means that $g$ has $n-1$ distinct roots in $L$. Since $g$ has degree $n-1$, this means that $g$ has all of its roots in $L$. Therefore, $g$ splits completely in $L$ and has all of its roots in $M$. This implies that $g$ is the minimal polynomial for $\alpha$ over $M$. Since $g$ has degree $n-1$, we have $[M(\alpha):M] = n-1$.
Now, we have $[M: