It is given that $p = 2q+1$, and $p$ is a prime number. Prove that 
$$(q!)^2 +(-1)^q$$ is a multiple of $p$. 

Attempt:

$q = \frac{p-1}{2}$

\begin{align}
\therefore (q!)^2 +(-1)^q &= \frac{1}{4} [(p-1)!]^2 +(-1)^q \\ &=\frac{1}{4} \left[\frac{p!}{p}\right]^2 +(-1)^q
\end{align}

The first part is clearly divisible by $p$. I dont know what to do with the $-1$.