ProofOfFLT

If $s,t\in\mathbb Z$ are such that $sn\equiv tn\pmod p$, then $s\equiv t\pmod n$ since $p$ and $n$ are coprime. Thus, we see that $n$, $2n$, $3n$, $\dots$, $(p-1)n$ are all distinct modulo $p$. Thus $$n^{p-1}\cdot (p-1)!=n\cdot 2n\cdot 3n\cdot \dots\cdot (p-1)n\equiv (p-1)!\pmod p,$$ and so $n^{p-1}\equiv 1\pmod p$ as required. $\Box$