Since you are actually studying for a test, I guess it wouldn't hurt to know this. You can say a little bit more about the group $G$. In general, if $p$ is a prime number and $G$ is a group of order $p^2$, then $G \cong \mathbb{Z}/p^2 \mathbb{Z} \quad \text{or} \quad G \cong \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$
Thus in particular every group of order $p^2$ is abelian and in your case, since you assume that the group is not cyclic, you have $G \cong \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$ so in this case it is immediate (I hope =P) that every element $a \in G$ satisfies $a^p = e$. You can see a short proof of this here.