How to show that $2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}$ 
Show that $2730\mid n^{13}-n,\;\;\forall n\in\mathbb{N}$

I tried, $2730=13\cdot5\cdot7\cdot3\cdot2$
We have $13\mid n^{13}-n$, by Fermat's Little Theorem.
We have $2\mid n^{13}-n$, by if $n$ even then $n^{13}-n$ too is even; if $n$ is odd $n^{13}-n$ is odd.
And the numbers $5$ and $7$, how to proceed?

HINT:
$$n^{13} \equiv n^5 \cdot n^5 \cdot n^3 \equiv n \cdot n \cdot n^3 \equiv n^5 \equiv n \pmod 5$$
$$n^{13} \equiv n^6 \cdot n^7 \equiv n \pmod 7$$
Also you've missed $3$ as prime factor. But that should be easy.
