Do you divide me inductively?

Fermat's Little Theorem tells us that $n^7\equiv n\pmod {\boxed{7}}$ .

$n^7-n=n(n^3-1)(n^3+1)=n(n-1)(n+1)\cdot (n^2+n+1)(n^2-n+1)$

$n(n-1)(n+1)$ is a product of $3$ consecutive integers, hence it is divisible by $\boxed{6}$ .

Hence, $\forall n\in\mathbb Z$ , we have $n^7\equiv n\pmod {\boxed{42}}$ , since $7\cdot 6=42$ .

$2^7-2=126=42\cdot 3$ and $3^7-3=42\cdot 52$ , so $42$ is the largest possibility, because $\gcd(3,52)=1$ .