Fix a finite field $\mathbb{F}_{p^k}$. Since its multiplicative group $\mathbb{F}_{p^k}^\times$ is cyclic,the primitive $(p^k-1)^{th}$ roots of unity are precisely the generators of $\mathbb{F}_{p^k}^\times$ of which there are $\varphi(p^k-1)$, where $\varphi$ is the Euler totient function. This makes sense so far.
From this mathoverflow answer, the number of $n$-th roots equals $\gcd{(n,p^k-1)}$. I am struggling to understand why. Letting $g$ be a generator for $\mathbb{F}_{p^k}^\times$, the first step in justifying the claim is
For an element $x$ of the group $x^n=1$ holds iff $x=g^m$ with $nm$ divisble by $p^k-1$
Why is this the case? I'm probably missing something super basic.
As you said, the group $\mathbb{F}_q^*$ ($q=p^k$) is cyclic of order $q-1$, so every element is of the form $g^m$ for some generator $g$. Of course, $g$ has order $q-1$. If $x=g^m$ is an $n$th root of unity, that is the same as saying $x^n=g^{mn}=1$ which is the same as $q-1\mid mn$.