Let $f\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\geq 5$.
Let $L$ be the splitting of $f$ and let $\alpha\in L$ be a zero of $f$.
Claim If $[L:\mathbb{Q}]=n!$, then $\mathbb{Q}[\alpha] = \mathbb{Q}[\alpha^4]$.
Attempt
$[L:\mathbb{Q}]= n!$, and $L' \leq S_n$ implies $L'= S_n$.
$[\mathbb{Q}[\alpha]:\mathbb{Q}]=n$ since $f$ is irreducible.
$\mathbb{Q}[\alpha^4] \subset \mathbb{Q}[\alpha]\implies [\mathbb{Q}[\alpha^4]:\mathbb{Q}] \leq n$.
Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$.
If it is $2$, $\mathbb{Q}[\alpha^4]'=A_n$, which is a normal subgroup hence $\mathbb{Q}[\alpha^4]$ is a splitting field over $\mathbb{Q}$.
How can I claim it is not possible?
Notation If $\mathbb{Q} \subset K \subset L$, $\mathbb{Q}' = S_n$, $L'=e$ and $K'=\{s\in S_n: sk=k \text{ for all }k\in K\}$
Note that $[L:\mathbb{Q}(\alpha)] = (n-1)!$. Moreover it's the splitting field of the polynomial $\frac{f(x)}{x-\alpha} \in {Q}(\alpha)[x]$, which is a polynomial of degree $n-1$. This means that the Galois group of the extension is $S_{n-1}$.
As you've noted the Galois group of the extension $\mathbb{Q}(\alpha^4) \subset L$ is $A_n$. Also as $\mathbb{Q}(\alpha^4) \subset \mathbb{Q}(\alpha)$ we must have that the Galois group of $\mathbb{Q}(\alpha) \subset L$ is contained in the Galois group of $\mathbb{Q}(\alpha^4) \subset L$. In other words $S_{n-1} \le A_n$. However this is an obvious contradiction.
Another way is to consider the extension field $\mathbb{Q}(\alpha^2)$. With similar reasoning we get that $[\mathbb{Q}(\alpha^2):\mathbb{Q}] = 2$ or $n$. If it's $2$, then we must have that $\mathbb{Q}(\alpha^2) = \mathbb{Q}(\alpha^4)$. However $\mathbb{Q}(\alpha^2) \subseteq \mathbb{Q}(\alpha)$ is a normal extension, as the latter field is the splitting field of $x^2 - \alpha^2 \in \mathbb{Q}(\alpha^2)[x]$. As $A_n$, the Galois group of the extension $\mathbb{Q}(\alpha^2) \subset L$ is a simple group for $n \ge 5$ we must get that $\mathbb{Q}(\alpha^2) = \mathbb{Q}(\alpha)$.
If $[\mathbb{Q}(\alpha^2):\mathbb{Q}] = n$, then we have that $\mathbb{Q}(\alpha) = \mathbb{Q}(\alpha^2)$. Similalry $\mathbb{Q}(\alpha^4) \subseteq \mathbb{Q}(\alpha^2)$ is a normal extension, which would yield equality between the fields, as the Galois group, $A_n$ is simple.
Hence the proof.