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?

One Approach
If $k\mid n$, then $x^{k+1}-x\mid x^{n+1}-x$. Therefore,
$$
\begin{array}{}
13&\mid&n^{13}-n\\
7&\mid&n^7-n&\mid&n^{13}-n&\text{since }6\mid12\\
5&\mid&n^5-n&\mid&n^{13}-n&\text{since }4\mid12\\
3&\mid&n^3-n&\mid&n^{13}-n&\text{since }2\mid12\\
2&\mid&n^2-n&\mid&n^{13}-n&\text{since }1\mid12\\
\end{array}
$$
Since the factors are all relatively prime, we have that
$$
2730=2\cdot3\cdot5\cdot7\cdot 13\mid n^{13}-n
$$

Another Approach
Decomposing into a sum of combinatorial polynomials
$$
\begin{align}
n^{13}-n
&=2730\left[\vphantom{\binom{n}{2}}\right.3\binom{n}{2}+575\binom{n}{3}+22264\binom{n}{4}+330044\binom{n}{5}\\
&+2458368\binom{n}{6}+10551552\binom{n}{7}+28055808\binom{n}{8}+47786112\binom{n}{9}\\
&+52272000\binom{n}{10}+35544960\binom{n}{11}+13685760\binom{n}{12}+2280960\binom{n}{13}\left.\vphantom{\binom{n}{2}}\right]\\
\end{align}
$$
