If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$ 
If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$.

I don't know how to prove this statement.
$p=4m+3$, so $(2m+1)! \equiv \pm1\pmod p$
This is all I did.

$\left( \frac{p-1}{2}! \right)^2 = \prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}n=\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}-n \equiv\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=\frac{p+1}{2}}^{p-1}n=(p-1)! \equiv 1$
You can change $n$ for $-n$ because there is an even number of factors from $1$ to $\frac{n-1}{2}$ 
Therefore, $ \frac{p-1}{2}!$ is congruent to either $1$ or $-1$
