Show that $[(\frac{p-1}{2})!]^2+1\equiv0 \pmod p$

Question

Let p be a prime such that $p\equiv1\mod 4$. Show that $[(\frac{p-1}{2})!]^2+1\equiv0 \pmod p$

I know I have to use Wilson's Theorem, but not sure how.

Answer 1

Modulo $p$, we have that $$-1\equiv (p-1)! \equiv \left(\frac{p-1}{2}\right)! \times \frac{p+1}{2} \times \frac{p+3}{2} \times \cdots p-1 \equiv \left(\frac{p-1}{2}\right)! \times -\frac{p-1}{2} \times -\frac{p-3}{2} \times \cdots (-1) \equiv (-1)^{(p-1)/2} \left(\left(\frac{p-1}{2}\right)!\right)^2$$

The result follows since $p \equiv 1 \ \text{mod} \ 4$.