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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?

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    Yes, Wilson's theorem is overkill. Note that $p$ is odd. – 2017-02-16
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    Hint: $p+1$ is even. What can you say about $(p+1)/2$? (Wilson's theorem is a tempting start but no help as you discovered.) – 2017-02-16
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    Why the downvote? It's a good question and the OP made a fair try before asking. – 2017-02-16
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    If p is prime then p+1 is not prime. So it's factors are less than p. So it's factors are components of p!. The only issue is what if p+1 is the square of a prime. – 2017-02-16
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    .... which isn't a concern as p+1 is even and and greater than 4. Doh! – 2017-02-16
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    @Ethan My guess is the odd symbols, combined with the fact that he knew enough TeX to use it in the title. Despite these oddities, I upvoted. – 2017-02-16
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    Why on earth did you use these symbols? – 2017-02-16

4 Answers 4

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$p+1$ 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.

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    And 9|8!. As long as p+1 isn't a square of a prime it has two factors less less than p. If it is a square of a prime then the prime and twice the prime are less than p so it's okay. Of course being even is even easier... but this is still good. – 2017-02-16
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    @jonas corrected. – 2017-02-16
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Let $p+1=2q$, then $1\leq q= \dfrac{p+1}{2}

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 $(p-i)=q$

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    Why work so hard? $2$ and $q$ are different and both less than $p$ so obviously both divide $p!$. (You could have waited a bit to give the OP a chance to use the hint.) – 2017-02-16
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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.

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If $p>3$, then $p+1 = 2q$, with $2