((p − 1)/2)! modulo a prime p ≡ 3 (mod 4)

Pairing. Wilson's theorem: $(p-1)!\equiv-1\pmod p$. Write
$$(p-1)!=\prod_{k=1}^m k\cdot\prod_{k=1}^m(p-k)\equiv\prod_{k=1}^mk\cdot\prod_{k=1}^m(-k)=(-1)^m(m!)^2\pmod p.$$

Parity of $m$. Since $p\equiv3\pmod4$, $m=\frac{p-1}2$ is odd, so $(-1)^m=-1$.

Conclusion. $-1\equiv-(m!)^2$, i.e. $(m!)^2\equiv1\pmod p$. Since $p$ is prime, $m!\equiv\pm1\pmod p$.

Check. $p=7$: $3!=6\equiv-1$. $p=11$: $5!=120\equiv-1$. $p=23$: $11!\equiv1\pmod{23}$.

Remark. For $p\equiv1\pmod4$ the same computation gives $(m!)^2\equiv-1$, which is the key to writing such a prime as a sum of two squares.

Lesson. Wilson's theorem plus the pairing $k\leftrightarrow p-k$ turns $(p-1)!$ into a square (up to sign) of half the product.