Proving that a group of order $p^nq$ for primes $p$ and $q$ is not simple. 
Prove that a group of order $p^nq$ for primes $p$ and $q$ is not simple.

I've been able to prove the theorem holds for $p=q$ and $p>q$. If $p<q$ the best I've been able to do is use Sylow to show: $$p^n+p^{n-1}-1\leq q$$
Yet I seem to be stuck. I would appreciate any help.

Let $P$ be a $p$-Sylow subgroup of $G$. Suppose $P$ is not normal.
Consider $n_p$, the number of $p$-Sylow subgroups conjugate to $P$. By Sylow, $n_p | q$ so since $n_p \neq 1$ it must equal $q$.
Note that the set of $p$-Sylow subgroups is exactly the set of subgroups conjugate to $P$. All these groups have the same order, so are all $p$-Sylow subgroups, and all $p$-Sylow subgroups are conjugate. Furthermore, the intersection of the subgroups conjugate to $P$ is normal. (See Intersection of conjugate subgroups is normal.) So, exploring the possibility that $G$ is simple, suppose that intersection is trivial.
Then the total number of non-trivial elements in all $p$-Sylow subgroups is $(p^n-1)q$. That leaves exactly $p^nq - (p^n-1)q - 1 = q-1$ non-trivial elements among all $q$-Sylow subgroups.
Each $q$-Sylow subgroup has order $q$, so it has $q-1$ non-trivial elements, so there is only one such subgroup which is therefore normal.
To recap: either the $p$-Sylow subgroup is normal, the intersection of all $p$-Sylow subgroups is non-trivial and normal, or the $q$-Sylow subgroup is normal.
