I want to show that there's no group of order $63$ which is simple and would like to know if my simple reasoning is correct. I am irritated because this is an exercise which is supposed to be harder.
So we have $|G|=63=3^27$. By the second Sylow theorem we know that the number of $7$-Sylow-subgroups $m_7 | 9$ and $m_7 \equiv 1 \mod{7}$. So there's only one $7$-Sylow-subgroup. This a normal subgroup because Sylow-subgrups are conjugate to each other. So $G$ isn't simple.
Using the Burnside theorem we can proof that if $G$ is not abelian group and $|G|=p^aq^b$ where $2\leq a+b$ and $p,q$ primes, so the problem is done if $G$ is not abelian, the proof of Burnside is not easy and use theory of representation finite group, look for example bejamin-steinber. I hope to be helpful. :)