If $|G|>1$ is not prime, there exists a subgroup of $G$ which is not $G$ or $e$.

If $|G|>1$ is not prime, say $|G|=ab$ with $a,b>1$. Select $g\in G\setminus\{1\}$. If $\langle g\rangle \ne G$, we are done. Otherwise, the order of $g$ is $ab$ and we can consider $\langle g^a\rangle$ instead, which has order $b$.

Looking at the subgroup $H$ generated by an element $g\ne e$ is a good idea. Perhaps this group is a proper subgroup. Then we are finished.

If $H$ is all of $G$, let $|G|=ab$ where neither $a$ nor $b$ is $1$, and consider the group generated by $g^a$. That seems to be what you were thinking of, but one needs to prove that under these conditions the group generated by $g^a$ is indeed a proper subgroup.

If instead we consider the subgroup generated by $g^n$, where we only specify that $1\lt n\lt |G|$, then this subgroup is often all of $G$. In fact it is all of $G$ precisely if $|G|$ and $n$ are relatively prime.

Hint:Let $o(G)=mn$ then one subgroup is $<g^m>$.