# Galois group of order $n!$

Suppose $$f\in\mathbb{Q}[x]$$ has degree $$n$$, and let $$K$$ be the splitting field of $$f$$ over $$\mathbb{Q}$$. Suppose the Galois group of $$K/\mathbb{Q}$$ is $$S_n$$ with $$n\geq 3$$. It is not hard to show that $$f$$ is irreducible and that for any root $$\alpha$$ of $$f(x)$$ the only automorphism of $$\mathbb{Q}(\alpha)$$ is the identity. Now, I am trying to understand why if $$n\geq 4$$ then $$\alpha^n\not\in\mathbb{Q}$$.

If $$\alpha\in\mathbb{Q}$$ then $$\alpha$$ satisfies a polynomial $$x^n-a/b$$, where $$a/b\in\mathbb{Q}$$ and we have that $$f(x)$$ divides this polynomial as it is the minimal polynomial for $$\alpha$$. However, how should I proceed to arrive at a contradiction?

One way to proceed is the following: Not only is the only automorphism of $$\mathbb Q(\alpha)$$ the identity, the only automorphism of $$\mathbb Q(\alpha,\beta)$$ for $$\alpha,\beta$$ roots of $$f(x)$$ are the identity and the one swapping $$\alpha$$ and $$\beta$$.
On the other hand, if the roots were of the form $$\alpha\zeta_n^i$$ for $$\zeta_n$$ a $$n$$-th root of unity and $$1\leq i \leq n$$ (as they would have to be if $$\alpha^n \in \mathbb Q$$), then $$\mathbb Q(\alpha,\alpha\zeta_n)$$ is Galois and contains all the roots and so has many automorphisms (for $$n\geq 3)$$.