If $p^k|n!$ then $p!^k|n!$

Induction on $n$: If $n=mp+r$ with $0\le r<p$, there are $\frac{n!}{p!^mr!m!}$ ways to partitoin $n$ objects into $m$ indistinguishable subsets of size $p$ each and a rest of size $r$. Note that $p^{k-m}\mid m!$ (because the multiples of $p$ among $1,2,\ldots, n$ are precisely $p,2p,,\ldots, mp$) and by induction hypothesis $p!^{k-m}\mid m!$. Hence $p!^n\mid p!^mm!\mid n!$.