If $p$ is an odd prime number, prove that $1^{2}3^{2}5^{2}\dotsb(p-4)^2(p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p$
I thought maybe I could use $p-k \equiv -k \pmod p$ to try and reduce the LHS, but I'm not sure how to account for the negative sign. Any help would be much appreciated.
We have that, modulo an odd prime $p$, $$\prod_{k=1}^{(p-1)/2}(p-2k)^2\equiv 4^{(p-1)/2}\prod_{k=1}^{(p-1)/2}k^2\equiv 2^{p-1}\prod_{k=1}^{(p-1)/2}k\cdot \prod_{k=1}^{(p-1)/2} (k-p)\\\equiv (-1)^{(p-1)/2}\prod_{k=1}^{(p-1)/2}k\cdot \prod_{j=(p+1)/2}^{p-1}j\equiv (-1)^{(p-1)/2}\prod_{k=1}^{p-1}k\equiv (-1)^{(p-1)/2}$$ where we used $2^{p-1}\equiv 1$ (Fermat's Little Theorem) and $\prod_{k=1}^{p-1}k\equiv 1$ (Wilson's Theorem).
