Show if $p > 3$ is prime, then $(p + 1) | p!$ I know I must show that p! $\equiv$ 0 mod p + 1. I am attempting to use Wilson's theorem. 
p! $\equiv$ p(p - 1)! $\equiv$ p $\cdot$ - 1 mod p. Since p $\equiv$ 0 mod p, then p! $\equiv$ 0 mod p...but this is obvious since p divides p. So this argument seems to be a dead end.
I tried the following: 
p must be odd since p is a prime larger than 3. Thus, p + 1 is even. Since p + 1 is even it must be the product of 2 and an integer n such that n < p + 1. Now, p! = p(p - 1) $\cdot\cdot\cdot$ n $\cdot\cdot\cdot$ 2 $\cdot$ 1. Thus p + 1 = 2n | p!
Is this a rigorous enough argument? 
 A: $p+1$ is an even number greater than 4, thus a product of at least one pair of distinct whole numbers smaller than itself.  Such a product divides $p!$ by construction.
A: Let $p+1=2q$, then $1\leq q= \dfrac{p+1}{2}<p$
Clearly, for all $p>3$, $p!\equiv0 \pmod 8$.
Since $p!=p(p-1)(p-2)\cdots2\cdot 1$ .  Then, for some $1\leq i\leq \dfrac{p+1}{2}$,  we must have $(p-i)=q$
A: If $p>3$, then $p+1 = 2q$, with $2<q<p$, so $p!$ evidently has factors of both. 
A: Another way: it is enough to show that if $q$ is a prime number, and $q^n$ divides $p+1$, then $q^n$ divides $p!$.
There are two cases: either $q^n = p+1$, or $q^n$ is a proper divisor of $p+1$.
In the first case, $n$ cannot be $1$, since otherwise $p$ and $p+1$ are both prime, which is contrary to the hypothesis.  If $n = 2$, then $p+1 = q^2$, and so $p = (q+1)(q-1)$, which is impossible unless $q = 2$, which implies that $p = 3$, contrary to the hypothesis.  So $n$ must be $\geq 3$, and so $nq < q^n = p+1$, which means that $nq \leq p$.  Then $q, 2q, 3q, ... , nq$, and hence their product, $n!q^n$, divide $p!$.  In particular, $q^n$ divides $p!$.  
For the second case, suppose that $q^n$ is a proper divisor of $p+1$.  Then again, $nq \leq q^n \leq p$, so the same argument at the end of the previous paragraph applies.
