Prove that if $a^p=e$ where $p$ is a prime number, then $a$ has order $p$ ($a\ne e$).
Perhaps I should prove this using the contrapositive? Assume that $p$ is not the order of $a$, meaning there exists a smallest positive integer $n$, where $n<p$ and ord($a$)=$n$, i.e., $a^n=e$.
Is that a good direction to go? If so, where do I go from there?
Hint:
Show that if $a^n = e$ , then $\mathrm{ord}(a) | n$. From that, it is clear that in your case, $p$ must be the order of $a$.