I am not sure how to approach the following problem:
Show that if $|G|=n$, then $|\mathrm{Aut}(G)|$ divides $(n-1)!$
All I can think of so far is that clearly $|\mathrm{Aut}(G)| \le |\mathrm{Sym}(G \setminus \{e\})| = (n-1)!$, but this doesn't give any suggestion as to why $|\mathrm{Aut}(G)|$ divides $(n-1)!$
Thank you.