Show that $30 \mid (n^9 - n)$ I am trying to show that $30 \mid (n^9 - n)$. I thought about using induction but I'm stuck at the induction step.
Base Case: $n = 1 \implies 1^ 9 - 1 = 0$ and $30 \mid 0$.
Induction Step: Assuming $30 \mid k^9 - k$ for some $k \in \mathbb{N}$, we have $(k+1)^9 - (k+1) = [9k^8 + 36k^7 + 84k^6 + 126k^5 + 126k^4 + 84k^3 + 36k^2 + 9k] - (k+1)$.
However I'm not sure where to go from here.

And here are the congruences requested to complement the answer by Simon S
$$\begin{array}{c|ccccc}
n&n-1&n+1&n^2+1&n^4+1\\
\hline
0&4&1&1&1\\
1&0&2&2&2\\
2&1&3&0&4\\
3&2&4&0&2\\
4&3&0&2&2
\end{array}$$
And one can see that one of these factors is always $\equiv 0\pmod{5}$
