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

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.

Like user99680,

Using Fermat's Little Theorem $p|(n^p-n)$ where $n$ is any integer and $p$ is any prime

$\displaystyle n^{13}-n=n(n^{12}-1)=n\left((n^6)^2-1\right)=n(n^6-1)(n^6+1)=(n^7-n)(n^6+1)$ which is divisible by $\displaystyle n^7-n$ which is always divisible by $7$ for all integer $n$

Similarly, $\displaystyle n^{13}-n=n(n^{12}-1)=n\left((n^4)^3-1\right)$ $\displaystyle=n(n^4-1)(n^8+n^4+1)=(n^5-n)(n^8+n^4+1)$ which is divisible by $\displaystyle n^5-n$ which is always divisible by $5$ for all integer $n$

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 $$ 2730=2\cdot3\cdot5\cdot7\cdot 13\mid n^{13}-n $$

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Another Approach

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