We prove that the only solutions are $n\in \{ 2,3 \}$. If $n$ is composite, let $n=a\cdot b$ and $a > b$. We have that $a!=1\cdot 2 \cdot 3 \hdots b \cdot (b+1) \hdots a$ so we have $(a+1)!\equiv (a+2)! \equiv \hdots \equiv (a+(n-a))! \equiv 0$ so we must have $n-a=1$ or $n=2$. We need to handle the case when $a=b$ since this is weird. Let $n=x^2$. I prove $(x+1)! \equiv \hdots (x^2)! \equiv 0 \ ( mod \ n)$ since $x^2|(x+1)!$ because $(x+1)!=(x-1)! \cdot x \cdot (x+1)$ but obviously $x|(x-1)!$ unless $x$ is prime, but we'll handle this too. So now we need to check when $x=p$. By Wilsons we have $(p-1)! \equiv -1 \ ( mod \ n )$ and we have $(p-2)! \equiv 1 \ (mod \ n)$ but this can't happen since $1! \equiv (p-2)!$ and basically we check $n=3$ and this is also a solution.