Let $K$ be a field of characteristic zero and consider a finite extension of the form $K(\alpha)$. This is often the setting in the first step of a radical extension.
If we know that $\alpha^n \in K$ for some minimal $n$, does it follow that $x^n - \alpha^n$ is the minimal polynomial of $\alpha$? This seems to be often assumed in the course I am learning but I can't see an immediate way of telling this is true. For all I know, it could be possible that for some smaller $k$, $\alpha ^k \notin K$ but you can use the rest of the terms in the polynomial in order to get a zero.
Would the situation be any better if we assume $K = \mathbb{Q}$?
EDIT: to not allow certain trivial cases, assume that $1 \neq \alpha^k \in \mathbb{Q}$
The assertation is false, from the fact that $\alpha^n \in K$ you know that it must be a root of $f=x^n-\alpha^n$, so it's minimal polynomial must divide $f$. If as $f$ has a root in $K$ we write $f=(x-a)g(x)$ and $\alpha \notin K$ so $\alpha \neq a$ and it's minimal polynomial must divide $g$ hence it's not $f$.
For e.g. take $K=\Bbb R$ and $\zeta = 2e^{\frac{2\pi i}{3}}$ then the polynomial $x^3-8 = (x-2^{1/3})(x-\zeta)(x-\bar{\zeta})$ hence (and this is easy to check) $\zeta$'s minimal polynomial is $x^2+2x+4$