Let $n=p^mr$ where $p$ is prime and $r\in\mathbb{Z_{>1}}$ such that $p\nmid r$. If $G$ is a simple group of order $n$ then $p^m\mid(r-1)!$ Question: Let $n=p^mr$ where $p$ is prime and $r\in\mathbb{Z^{>1}}$ such that $p\nmid r$.  If $G$ is a simple group of order $n$ then $p^m\mid(r-1)!$.
My Ideas:  I feel like this should be a rather straightforward argument but I keep getting tripped up.  Since $G$ is simple, the only normal subgroups of $G$ are $\langle1\rangle$ and $G$. So, we have a homomorphism $\phi:G\rightarrow S_n$, where $n=1$ or $n=p^mr$. Since $G$ is simple, $\ker\phi=\langle1\rangle$.  If $n=1$, then we would violate the assumption that $r\in\mathbb{Z}_{>1}$.  .....and I am sort of stuck from here... maybe this isn't the best way to go about it?  Any help is greatly appreciated!  Thank you

You get a sharper bound by letting $G$ act on the conjugates of $P$ (the Sylow $p$-subgroups) instead of on the cosets; then there are $n_p = \frac{|G|}{|N_G(P)|} = \frac{r}{s}$ of these, where $s = \frac{|N_G(P)|}{|P|}$, and you get that $p^m r$ divides $n_p!$. If $r \not \equiv 1 \bmod p$ you're guaranteed that $n_p < r$ by the Sylow theorems so this bound will be strictly sharper in that case.
For example, if $|G| = 300$ and $p = 5$ then $r = 12$ and the given bound gives $25 \mid 11!$, which is true, but the sharper bound gives that $n_p = 6$ (it can't be equal to $1$ if $G$ is simple) so $25 \mid 5!$, which is a contradiction. So there's no simple group of order $300$.
