TITLE: Show that $\alpha^n\notin k$ for $n\ge 4$ where $\alpha\in\bar{k}$ and $[K:k]=n!.$
QUESTION [7 upvotes]: Let $k$ be a field, let $f(X) \in k[X]$ be a separable polynomial of degree $n$ whose Galois group is isomorphic to $S_n$ , and let $\alpha$ be a root of $f(X)$ in some algebraic closure $k$.
(a) Show that $f(X)$ is irreducible.
(b) Show that $Aut_k(k(\alpha)) = \{\mathrm{id}\}$ if $n \ge 3.$
(c) Show that $\alpha^n \notin k$ if $ n\ge 4.$

I know what to do for (a) and (b):
(a) Let $G$ be the Galois group of $f$. For the sake of argument, suppose $f(x)$ is reducible. Then $f$ can be written as a product of distinct irreducible factors (since $k[x]$ is a UFD). Let $\alpha_i, \alpha_j$ be distinct roots of two such irreducible factors. But then any $\sigma\in G$ must send $\alpha_i$ to another root of its minimal polynomial. In particular, we can't have $\sigma(\alpha_i) = \alpha_j$, hence $G$ is not isomorphic to a transitive subgroup of $S_n$. But $S_n$ is a transitive subgroup of itself, a contradiction.
(b) It suffices to show that $\alpha$ is the only root of $f(x)$ in $k(\alpha)$. Suppose there is another root $\beta\in k(\alpha)$. Then over $k(\alpha)$,
$$f(x) = (x-\alpha)(x-\beta)g(x),$$
where $\deg g(x) = n-2.$ Let $K$ be the splitting field of $f$ over $k$. Then since $G\cong S_n,$ we have $[K:k]=n!.$ And since $\deg g = n-2,$ we have $[K:k(\alpha)]\le (n-2)!,$ and $[k(\alpha):k]=n$ by the irreducibility of $f$ over $k$. But then
$$[K:k] = [K:k(\alpha)][k(\alpha):k]\le (n-2)!\cdot n<n!,$$
contradicting that $G\cong S_n.$
Now how do we attack (c)?

REPLY [6 votes]: Suppose $\alpha^{n} \in k$. Then $\alpha$ is a root of $g(X) = X^{n} - \alpha^{n}$, so $f \mid g$; since both are monic and of the same degree, we get $f = g$. The roots of $g(X)$ are of the form $\xi^{k}\alpha$, where $\xi$ is a primitive $n$th root of unity, so the splitting field $K$ of $g$ is given by $k(\xi, \alpha)$. By part $b$, the splitting field $K$ must have degree $n!$ over $k$; but $[k(\xi, \alpha):k] \leqslant [k(\xi):k]\cdot [k(\alpha):k] = \varphi(n) \cdot n \leqslant n(n-1)$, where $\varphi$ is the Euler totient function. If $n \geqslant 4$, then $n! > n(n-1)$, a contradiction. 
It's maybe worth noting that we can't improve the bound $n \geqslant 4$. Taking $k = \mathbb{Q}$, the polynomial $f(X) = X^{3}-2$ has
Galois group $S_{3}$, and the roots are $\sqrt[3]{2}, \xi\sqrt[3]{2}, \xi^{2}\sqrt[3]{2}$, where $\xi$ is a primitive
third root of unity.