Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / C / 1 / iii / Solution

Combine the preceding congruences: $a^p\equiv a$ and $(p-1)!\equiv -1(\mathrm{mod}p)$. Therefore $a^p+(p-1)!a\equiv a-a=0$, and $(p-1)!a^p+a\equiv -a+a=0$. This proves both stated divisibilities for every integer $a$, with no assumption that $a$ is coprime to $p$.