Prove that if $p \not \mid n$ then
$$ n^{k(p−1)} \equiv 1 \pmod {p}$$
where $p$ is prime and $k$ is a natural number.
Since $p \not \mid n$ we can use the FlT
$$ n^{p-1} \equiv 1 \pmod p $$
Using $a\equiv b \pmod n \implies a^k \equiv b^k \pmod n$,
$$ (n^{p-1})^k \equiv 1^k \pmod p $$
which simplifies to our desired conclusion.
$$ n^{k(p-1)} \equiv 1 \pmod p $$