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?

Cute corollary to FLT.
If $p,q $ are primes and $p-1=m|q-1=k$ then
$p|n^p -n=n (n^m-1)|n (n^m-1)(n^{k-m} + n^{k - 2m}+...+n^m+1)=n (n^k-1)=n^q-n $.
So as $1,2,3,4,6,12$ all divide $12$, it follows $2,3,5,7,13$ all divide $n^{13}- n $.
For me personally it was hardest to see that 5 did but $n^{13}-n =  (n^8 + n^4 + 1)(n^5- n) $ so ... it does.
