How prove $p!|(a+1)$ if $p!|a^p+1$

let $p$ is odd prime numbers, and there exsit postive ineteger $a$ such $$p!|a^p+1$$

show that $$p!|(a+1)$$

I think maybe can use $$a^p+1=(a+1)(a^{p-1}-a^{p-2}+\cdots-a+1)$$ then I can't Continue

• $p$ does not divide $\phi(p!) = \phi((p-1)!).(p-1)$, hence ${\mathbb{Z}}/p!^\times \rightarrow {\mathbb{Z}}/p!^\times,~ x \mapsto x^p$ is injective. Thus $a^p = -1 \mod p! \Rightarrow a = -1 \mod p!$ – user10676 May 30 '14 at 14:53

Pick any prime $q<p$, and suppose $q^k \|p!$. Then $p! \mid a^p+1$ so $q^k \mid a^p+1$. Now $a^p \equiv -1 \pmod{q^k}$ and $a^{2p} \equiv 1 \pmod{q^k}$. Let $d$ be the order of $a \pmod{q^k}$. Then we have $d \nmid p, d\mid 2p$. Also by Euler's totient theorem $a^{q^{k-1}(q-1)} \equiv 1 \pmod{q^k}$, so $d \mid q^{k-1}(q-1)$. Since $q<p$, we have $\gcd(q^{k-1}(q-1), p)=1$, so $p \nmid d$. Thus $d=2$, so $q^k \mid a+1$. This shows $(p-1)! \mid a+1$.
Finally $p \mid a^p+1$ so by Fermat's little theorem, $p \mid a+1$. Thus combining gives $p! \mid a+1$.