Prove that $p^2 | (p!+p)$ for any prime $p.$

Question

Prove that $p^2 | (p!+p)$ for any prime $p.$

So far I have: If $p$ is prime, then $p$ divides $[(p-1)!+1].$ Therefore $(p-1)!+1\equiv 0 \operatorname{mod}p.$ Where do I go from there?

Answer 1

We have $(p-1)!=px-1$ where $x\in \Bbb Z.$ Therefore $p!+p=p\cdot (p-1)!+p=p(px-1)+p=p^2x.$

Answer 2

$(p-1)!+1=0 \pmod p$ is Wilson's theorem.

Answer 3

multiplying both sides by $ p $

to get

$$ p(p-1)! + p = p( p n) $$

where $n$ is natural number

hence

$$ p! + p = p^2 n $$