elementary number theory - 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.

Answer

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 $\equiv 0\pmod{5}$