Proof of Fermat's little theorem in the form a^p = a

Proof

We argue by cases on whether $p$ divides $a$.

1. Let $p$ be a prime number and let $a$ be an integer.
2. Case $p \mid a$. Write $a = p s$ with an integer $s$. Since $p \ge 2$, we have $a^p - a = p s (a^{p-1} - 1)$.
3. So $p$ divides $a^p - a$, and $a^p \equiv a \pmod p$ by the definition of congruence.
4. Case $p \nmid a$. By Fermat's little theorem, $a^{p-1} \equiv 1 \pmod p$.
5. Since $a \equiv a \pmod p$, compatibility with multiplication gives $a^{p-1} \cdot a \equiv 1 \cdot a \pmod p$, that is, $a^p \equiv a \pmod p$.
6. In both cases, $a^p \equiv a \pmod p$.