Wilson's Theorem states:
$$ (p - 1)! ≡ -1 \mod p $$
If we expand:
$$ (1)(2) \ldots \left( \frac{p-1}{2} \right) \left( \frac{p+1}{2} \right) \ldots(p-2)(p - 1) ≡ -1 \mod p $$
Since \(p - x ≡ -x \mod p\):
$$ (1)(2) \ldots \left( \frac{p-1}{2} \right) \left( \frac{p+1}{2} \right) \ldots(-2)(-1) ≡ -1 \mod p $$
Let's take the negative sign out:
$$ (1)(2) \ldots \left( \frac{p-1}{2} \right) \left( \frac{p-1}{2} \right) \ldots(2)(1) \times \left( (-1)^{\frac{p-1}{2}} \right) ≡ -1 \mod p $$
The second half is the same size as the first. If \(p=4k+3\), then \(\frac{p-1}{2} = 2k+1\), which means:
$$ \left ( (1)(2) \ldots \left( \frac{p-1}{2} \right) \right)^2 \times (-1) ≡ -1 \mod p $$
If we multiply both sides by -1:
$$ \left ( \left( \frac{p-1}{2} \right) ! \right)^2 ≡ 1 ≡ 1^2 \mod p $$
According this lemma:
$$ \left( \frac{p-1}{2} \right) ! ≡ ±1 \mod p $$