Show that $x^{p^m} - x$ divides $x^{p^n} - x$ if and only if $m$ divides $n$. Working towards the complete classification of finite fields in our algebra class, some final book-keeping involved proving the above (for a prime $p$, $n \geq 1$). I've tried comparing the factorised forms: $$x^{p^n} - x = \prod_{a \in \mathbb{F_{p^n}}} (x-a)$$ but nothing is jumping out at me - I'm not even convinced the statement is true myself, yet. Any hints or intuition on why this is true or how to proceed would be appreciated. Thanks

Both are divisible by $x$ so $$x^{p^m}-x\mid x^{p^n}-x\qquad\iff\qquad x^{p^m-1}-1\mid x^{p^n-1}-1.$$ By this question we have $x^a-1\mid x^b-1$ if and only if $a\mid b$, so applying this twice shows that $$x^{p^m-1}-1\mid x^{p^n-1}-1 \qquad\iff\qquad p^m-1\mid p^n-1 \qquad\iff\qquad m\mid n.$$