The following is an exercise in "Certain Number-Theoretic Episodes In Algebra" by R. Sivaramakrishnan (in page 571).
(Jacobi) Let $p$ be a prime of the form $4k+3$. Show that $$ (2k+1)! + (-1)^{\eta} \equiv 0 \ (\operatorname{mod} p)$$ where $\eta$ is the number of quadratic non-residues of $p$ which are less than $p/2$.
I am interested in finding a reference for this particular result. If the proof is essentially self-contained, feel free to post it in the answer box. :)
By Wilson's theorem, we know $(4k+2)!\equiv -1 \pmod p$.
But we also have $(4k+2)!= (2k+1)! (p-1)(p-2)\dots(p-(2k+1)) \equiv -(2k+1)!^2 \pmod p$.
Therefore, $(2k+1)! \equiv \pm 1 \pmod p$.
But, $\pm 1 = \left( \frac{\pm1}p\right)$, since $-1$ is a non-residue for a prime of this form.
This gives $$(2k+1)! \equiv \prod_{i=1}^{2k+1} \left( \frac i p\right) = (-1)^{\eta} \pmod p.$$
Which obviously gives you the desired representation as $(-1)$ to the number of non-residues below $p/2$. (However, there is a wrong sign in your statement. The case $k=1$ supports my result.)