Solution:
Suppose that $p \mid a^{p}$. Since $p \leq q$, it follows that $a^{p} \mid a^{q}$, which implies that $p \mid a^{q}$.

Now, suppose that $p \mid a^{q}$, and, on the contrary, $p \nmid a^{p}$. Then there is a prime number $r$ and a positive integer $n$ such that $r^{n} \mid p$ (which implies that $r^{n} \leq p$) and $r^{n} \nmid a^{p}$. Since $p \mid a^{q}$, it follows that $r \mid a$, and so $r^{n} \mid a^{n}$. This means that $p < n$, which gives the following contradiction:
$$2^{p} \leq r^{p} < r^{n} \leq p$$
Therefore, $a^{p}$ must also be divisible by $p$.