Note that in $\{1,2,...,p^{2}-1\}$ there are $p-1$ multiples of $p$ ($p,2p,...,\left(p-1\right)p$) and none of these are multiples of $p^{2}$. Thus $p^{p}\nmid n!$ when $n<p^{2}$. But when $n=p^{2}$ we have that $p^{p+1}\mid n!$ (we have $1$ factor of $p$ from the $p-1$ multiples before and then $2$ factors of $p$ from $p^{2}$). Thus for $n\geq p^{2}$ we have that $p^{p+1}\mid n!$. Hence it is never the case that both $p^{p}\mid n!$ and $p^{p+1}\nmid n!$ so we are done.