Let $p$ be a prime and $\mathbb{F}_{p^k}$ be a finite field with (proper) subfield $\mathbb{F}_{p^a}$.
How many elements of $\mathbb{F}_{p^a}$ can be written as $x^{p^a-1}$ for some $x \in \mathbb{F}_{p^k}$?
From some small cases done computationally, I think it might depend on the prime $p$, but have no idea how to prove it. Any advice appreciated!
Note that $a\mid k$, and $p^a-1\mid p^k-1$. Consider nonzero elements. Note that $\mathbb{F}_{p^k}^*$ is cyclic of order $p^k-1$, say generated by $y$. We have $\mathbb{F}_{p^a}^*$ is a subgroup of order $p^a-1$, so is generated by $y^{\frac{p^k-1}{p^a-1}}$. Also, the image of $x\mapsto x^{p^a-1}$ is the subgroup generated by $y^{p^a-1}$. Hence, the number we want is the size of the intersection of the two subgroups, that is, the size of the subgroup generated by $y^{\operatorname{lcm} (p^a-1, \frac{p^k-1}{p^a-1})}$. This is $$ \frac{p^k-1} {\operatorname{lcm} \left(p^a-1, \frac{p^k-1}{p^a-1}\right)} = \operatorname{gcd}\left(p^a-1,\frac{p^k-1}{p^a-1}\right) $$ We can add one to count zero.