Proof Using Wilson's Theorem

I have been asked to prove the following for a prime $p$: $$ \frac{(np)!}{n!p^n} \equiv (-1)^n \pmod p $$ I know that Wilson's Theorem states that: $$ (p-1)! \equiv -1 \pmod p $$ but I cannot link the two! Please help.

We have

$$\begin{align*}\frac{(np)!}{n!p^n}&=\frac{1\cdot2\cdot3\cdots(np-1)\cdot np}{p\cdot2p\cdots np}\\ &=[1\cdot2\cdots (p-1)]\cdot\not p\cdot[(p+1)\cdot(p+2)\cdots(2p-1)]\cdot\not2\not p\cdots[((n-1)p+1)\cdots(np-1)]\cdot\not n\not p\end{align*}$$

By Wilson, the product of $p-1$ consecutive integers coprime to $p$ is $-1$ mod $p$. There are $n$ such products, so that makes $(-1)^n$.