Let $p$ be a prime number with $p \equiv 3 \pmod 4$. Prove that (${p-1} \over {2}$)! $\equiv (-1)^{n} \pmod p$

Let $p$ be a prime number with $p \equiv 3 \pmod 4$. Prove that (${p-1} \over {2}$)! $\equiv (-1)^{n} \pmod p$, where n is the number of positive integers less than $p \over 2$ that are quadratic nonresidues of p.

Pretty sure this involves Wilson's theorem, but I don't have many other ideas. Especially the "number of positive integers less than $p \over 2$ that are quadratic nonresidues of p" part.

By Wilson' s theorem $$\left[\frac{p-1}{2}!\right]^2\equiv 1 (\mod p)$$ Here the congruence $p\equiv 3\mod 4$ was used, but another consequence is that $-1$ is a quadratic non residue. Now for every $k$, $$k\equiv x^2\mod p$$ or $$k\equiv -x^2\mod p$$ according as $k$ is or is not a quadratic residue, and where we can assume that $1\leq x\leq \frac{p-1}{2}$. Further as $k$ runs through this domain, $x$ also achieves each values once. Thus we have that $$\frac{p-1}{2}!\equiv (-1)^n\left[\frac{p-1}{2}!\right]^2 \mod p$$