Let $p$ be the smallest prime divisor of $n$.

If $n\ne p^2$, then both $p$ and $n/p$ are less than $n$ and distinct. Thus $(n-1)!$ is divisible by $p(n/p)=n$.

If $n=p^2$, then $p > 2$ because $n > 4$. It follows that both $p$ and $2p$ are less than $n$. Therefore $(n-1)!$ is divisible by $p\cdot 2p=2p^2 = 2n$, hence by $n$.