 Solution

Since $n>1$, its factorial is even, so $(n!{)}^2+1$ is odd and its prime divisor $p$ is odd. Also $p$ cannot divide $n!$, since it would then divide1. Modulo $p$,Raise this equation to the integer power $(p-1)/2$. Fermat's little theorem givesAs $p$ is odd, $1$ and $-1$ are distinct residues. Thus the exponent is even, and