Computing $\gcd\{n^k - n^\ell : n \in \mathbb Z\}$

Let $r$ be a positive integer, and let $S_r$ be the set of positive integers $d$ such that the Euler's function $\phi(d)$ divides $r$. Let $M$ be the least common multiple of the numbers in $S_r$. If $a$ is the exponent of the highest power of a prime in the factorization of $M$ then, by Euler's theorem, $$\text{$n^{a+r}-n^a=n^a(n^r-1)$ is divisible by $M$.}$$

In your case: $r=6$, $S_r=\{2,3,4,6,7,9,14,18\}$, $M=252=2^2\cdot 3^2\cdot 7$ and $a=2$ which the statement above implies that $$\text{$n^{8}-n^2$ is divisible by $252$.}$$

Another example: if $r=12$ then $M=32760$, $a=3$ and $$\text{$n^{15}-n^3$ is divisible by $32760$.}$$