Let $\zeta_n := e^{2\pi i/n}$ and $K_n := \mathbb Q(\zeta_n) \cap \mathbb R$.
Suppose $\alpha \in K_n$ satisfies $\alpha^m \in \mathbb Q$ for some $m \geq 1$ such that $\alpha^j \notin \mathbb Q$ for all $1 \leq j < m$.
I am trying to show that $m \leq 2$.
Claim
$g(x) := x^m - \alpha^m$ is an irreducible polynomial in $\mathbb Q[x]$.
I could easily show that $K_n/\mathbb Q$ is a Galois extension where $\mathrm{Gal}(K_n/\mathbb Q)$ is abelian. Thus, $\mathbb Q(\alpha)/\mathbb Q$ must be a normal extension. Assuming the claim, we see that $g$ splits in $\mathbb Q(\alpha)$.
Since $\mathbb Q(\alpha) \subseteq K_n \subseteq \mathbb R$, all roots of $g(x) = x^m - \alpha^m$ must be real-valued. This happens if and only if $m \leq 2$.
As an attempt to prove the claim, I considered the minimal polynomial of $\alpha$ over $\mathbb Q$.
Let $f := \mathrm{minpoly}_{\alpha}(\mathbb Q)$. With the same argument as above, $\mathrm{deg}(f) \leq 2$ and all roots of $f$ must be real-valued.
If $\mathrm{deg}(f) = 1$, then $f(x) = x - \alpha$, which implies that $\alpha \in \mathbb Q$. In this case, $m=1$. Hence, assume that $\mathrm{deg}(f) = 2$.
Since $f$ has two real roots, and $f$ divides $g$, it follows that $g$ must have at least two real roots. By considering case by case, I could deduce that $\alpha^m > 0$ and $m$ is even.
This is where I am stuck at the moment. I would appreciate any hint for this problem.