$2(n-1)! \equiv -1 \mod n+2 \iff n+2$ is a prime Problem: Show that $2(n-1)! \equiv -1 \mod n+2 \iff n+2$ is a prime.

I know that Wilson's theorem states that $(n-1)! \equiv -1 \mod p $ for $p$ a prime, so that is the important thing to know with these type of problems.

I know that $-1 \equiv n+1 \mod n+2$ , 
and that if $n+2$ is prime, then $(n+1)!  = -1 \mod n+2$. So this is all I have for now, any hints appreciated.
As per the comments since I know that $2 (n-1)! \equiv (-1)(-2)(n-1)! \equiv (n)(n+1)(n-1)! \equiv (n+1)! \equiv -1 \mod (n+2)$, then $n+2$ is prime.
As per the comments since I know that $2 (n-1)! \equiv (-1)(-2)(n-1)! \equiv (n)(n+1)(n-1)! \equiv (n+1)! \equiv -1 \mod (n+2)$, then $n+2$ is prime.