If $p=4k+1$ then $p$ divides $n^2+1$ I am stuck in one step in the proof that if $p$ is congruent to $1 \bmod 4$, then $p\mid (n^2+1)$ for some $n$.
The proof uses Wilson's theorem, $(4k)!\equiv -1 \pmod p$.
The part I am stuck is where it is claimed that $(4k)!\equiv (2k)!^2 \bmod p$. Why is this so?

You group $\{1,p-1\},\{2,p-2\},...,\{2k,2k+1\}$ together, and the product of the first terms are equal to to product of second terms as there are even number of terms.
