Proving Fermat's little theorem

Permutation. The map $k\mapsto ak\bmod p$ is injective on $\{1,\ldots,p-1\}$ (if $ak\equiv a\ell$, then $k\equiv\ell$ since $a$ is invertible). So it permutes $\{1,\ldots,p-1\}$.

Product. $\prod_{k=1}^{p-1}(ak)\equiv\prod_{k=1}^{p-1}k\pmod p$, i.e. $a^{p-1}(p-1)!\equiv(p-1)!\pmod p$.

Cancel. Since $(p-1)!\not\equiv0\pmod p$ (all factors are nonzero), we get $a^{p-1}\equiv1\pmod p$.

Conclusion. Fermat's little theorem holds.