How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.

There are many equivalent ways of proving it.
First observe that $42$ divides a number iff $2,3$ and $7$ divides the number.
(Since $42 = 2 \times 3 \times 7$ and $\gcd(2,3) = \gcd(3,7) = \gcd(2,7) = 1$)
Divisibility by $2$:
Clearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.
Equivalently you could argue out that $2|(n^2-n)$ directly from Fermat's little Theorem. (This is an overkill of Fermat's Little Theorem.)
Divisibility by $3$:
$n^7-n = n(n^6-1) = n(n^2-1)(n^4+n^2+1)=n(n+1)(n-1)(n^4+n^2+1)$.
$3|n$ or $3|(n-1)$ or $3|(n+1)$ and hence $3|(n^7-n)$.
Equivalently you could argue out that $3|(n^3-n)$ directly from Fermat's little Theorem.
Divisibility by $7$:
Note that $n$ can be either $7k$ or $7k \pm 1$ or $7k \pm 2$ or $7k \pm 3$.
If $n=7k$ or $n=7k \pm 1$, we are again done since then $7|n$ or $7|(n+1)$ or $7|(n-1)$ and hence $7|(n^7-n)$.
If $n=7k \pm 2$, then $n^2 = (7k \pm 2)^2 = 7m + 4$ and $n^4 = (7m+4)^2 = 7l+2$.
Hence $n^4 + n^2 + 1 = 7l+2 + 7m + 4 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
If $n=7k \pm 3$, then $n^2 = (7k \pm 3)^2 = 7m + 2$ and $n^4 = (7m+2)^2 = 7l+4$.
Hence $n^4 + n^2 + 1 = 7l+4 + 7m + 2 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
Hence, $7|(n^7-n)$.
Equivalently you could argue out that $7|(n^7-n)$ directly from Fermat's little Theorem.
Therefore, we have that $2|(n^7-n)$ and $3|(n^7-n)$ and $7|(n^7-n)$, $\forall n \in \mathbb{N}$.
Hence, $42|(n^7-n)$, $\forall n \in \mathbb{N}$.
