Start by listing the first $p - 1$ positive multiples of $a$:
$$a, 2a, 3a, \ldots, (p - 1)a$$
Suppose that $ra$ and $sa$ are the same modulo $p$, then we have $r \equiv s \pmod p$, so the $p - 1$ multiples of a above are distinct and nonzero; that is, they must be congruent to $1, 2, 3, \ldots, p - 1$ in some order.
Multiply all these congruences together and we find
$$a \cdot 2a \cdot 3a \cdot \ldots \cdot (p - 1)a = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (p - 1) \pmod p$$
or better, $a(p - 1)(p - 1)! \equiv (p - 1)! \pmod p$. Divide both sides by $(p - 1)!$ to complete the proof.
Sometimes Fermat's little theorem is presented in the following form:
Corollary. Let $p$ be a prime and $a$ any integer, then $ap \equiv a \pmod p$.
Proof. The result is trival (both sides are zero) if $p$ divides $a$. If $p$ does not divide $a$, then we need only multiply the congruence in Fermat's little theorem by $a$ to complete the proof.