If $2p+1$ is a prime number, then $(p!)^2+(-1)^p$ is divisible by $2p+1$. I have a question which is as following:

If $2p+1$ is a prime number, then $(p!)^2+(-1)^p$ is divisible by $2p+1$.

Whenever I see 
questions involving factorial sign, I try to use Wilson theorem, but I don't know weather I can use it here or not, If yes then how??
Please help. 
 A: $p=2$ works. Suppose now $p\ge 3$. Then by Wilson's theorem
$$
(p!)^2= (-1)^p p!\cdot (-1)(-2)\cdots (-p)\equiv (-1)^p (2p)!\equiv (-1)^{p+1}\bmod{2p+1}.
$$
A: By Wilson's theorem since $2p+1$ is a prime number, $(2p)!+1$ is divisible by $2p+1$. Put $n$ for $2p+1$, then $(2p)!$ may be written as, $$1\cdot (n-1)\cdot 2 \cdot (n-2) 3\cdot(n-3) \cdots p(n-p)$$ if these factors supposed be multiplied out, it is obvious that we obtain $(-1)^p1^22^23^2\cdots p^2$ together with some multiple of $n$.   

Hence, $1+(-1)^p(p!)^2$ must be divisible by $n$ and therefore $(p!)^2 + (-1)^p$  must be divisible by $n$, i.e, $2p+1$. Hope it helps.
A: Since it is given that $(2p+1)$ is a prime number. So we can use Wilson theorem for $2p+1$ as:
$$ ((2p + 1)-1)! + 1\equiv 0 \mod(2p + 1) $$
$$(2p)!+1\equiv 0 \mod(2p + 1)$$
$$1\times2\times3\times4\times5\times6....(p)\times(p + 1)\times(p + 2)....(2p - 1)\times(2p) + 1\equiv 0 \mod(2p + 1)$$
$$p! [(2p+1-p)]\times  [2p+1-(p-1)]\times[2p+1-(p-2)]\times[2p+1-(1)]+1\equiv 0 \mod (2p+1)$$
$$p![M(2p + 1)+(-1)\times(p)\times(p-1)\times(p-2)....3\times2\times1] +1\equiv 0 \mod(2p + 1)$$
$$p![M(2p + 1)+(-1)^p {p!}] + 1\equiv 0 \mod(2p + 1)$$
$$[p!M(2p + 1)]+[(-1)^p {p!}^2] + 1\equiv 0 \mod(2p + 1)$$
Notice that $[p!M(2p + 1)]$ is already divisible by $(2p + 1)$, So,
$$[(-1)^p {p!}^2] + 1\equiv 0 \mod(2p + 1)$$
Which was needed to be proved.
:) :) :) :)
