Prove that if an element $a$ of a group is of prime order $p$ then either $a^m=e$ or $a^m$ has order $p$

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So for a bit of context I am working by myself through Arnold's "Abel's Theorem in Problems and Solutions" and I have been stuck on this problem for almost a week. I realise that it's elementary and I understand the other problems in that section, but for some unfathomable reason I just can't get past this one. It's incredibly frustrating when you can't understand something this simple, I feel like a complete dunce.

Arnold's proof uses the result from the previous problem, namely that if an element $a$ of the group has order $n$ then $a^m=e$ if and only if $m=nd$ where $d$ is any integer (0). This is basic and self-evident and I have no problem understanding it. For "my" problem he argues as follows:

$(a^m)^p=(a^p)^m=e$ (1), hence (see (0)) the order of the element $a^m$ must divide the number $p$ (2). Since $p$ is prime the statement follows (3).

I understand (1) by itself and why (3) follows from (2). But for the life of me I can't figure out how (2) follows from (1) and (0). I feel like someone is keeping my eyes shut with their hands and teasing me. Can anyone reformulate this in some other way?

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It's confusing because $a$ and $m$ have two different roles.

Rewrite $(0)$ as $b^p=e\iff $ord$(b) | p$.

Given $(1): (a^m)^p=e,$ take $b=a^m$ and therefore ord$(a^m)|p.$

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Hint: Recall that it suffices to prove that $\text{ord}(a^m)\text{ | }\text{ord}(a)$. What can you say about the subgroups generated by $a$ and $a^m$?