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.

Here's an alternative way of doing it:
Since $\phi(30)=8$, we get that $m^9 \equiv m \pmod{30}$ whenever $(m,30)=1$.
Since the only factors of $30$ are $2$,$3$, and $5$, we write:
$$ n=2^a \cdot 3^b \cdot 5^c \cdot m$$
Where $(m,30)=1$. Now:
$$ n^9 = 2^{9a} \cdot 3^{9b}\cdot 5^{9c} \cdot m^9 \equiv (2^9)^a \cdot (3^9)^b \cdot (5^9)^c \cdot m \pmod{30}$$
Now note: 
$$2^9 \equiv 2 \pmod{30}$$
$$3^9 \equiv 3 \pmod{30}$$
$$5^9 \equiv 5 \pmod{30}$$
So therefore: $$n^9 \equiv 2^a \cdot 3^b \cdot 5^c \cdot m \equiv n \pmod{30}$$
And therefore $30 \mid n^9-n$
