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?
 A: 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 $n$
A: 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
$$
\begin{align}
n^{13}-n
&=2730\left[\vphantom{\binom{n}{2}}\right.3\binom{n}{2}+575\binom{n}{3}+22264\binom{n}{4}+330044\binom{n}{5}\\
&+2458368\binom{n}{6}+10551552\binom{n}{7}+28055808\binom{n}{8}+47786112\binom{n}{9}\\
&+52272000\binom{n}{10}+35544960\binom{n}{11}+13685760\binom{n}{12}+2280960\binom{n}{13}\left.\vphantom{\binom{n}{2}}\right]\\
\end{align}
$$
A: Notice that $n^{13}-n =n(n^{12}-1)=n(n^6+1)(n^6-1)=n(n^6+1)(n^3+1)(n^3-1)=...$
A: Note that $$n^{13} \equiv n^5 \cdot n^5 \cdot n^3 \equiv n \cdot n \cdot n^3 \equiv n^5 \equiv n \pmod 5 \equiv n^{13} \equiv n^6 \cdot n^7 \equiv n \pmod 7.$$
A: $\, n =  2730 = 2\cdot 3\cdot 5\cdot 7\cdot 13 = \,$ product of all primes $\rm \,p\,$ such that $\rm \ \color{#c00}{p\!-\!1\mid 13\!-\!1}.\,$ Now apply 
Theorem $\ $  For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$ 
$\qquad\rm n\:|\:a^{\large e}-a\:$ for all $\rm\:a\:\iff  n\:$ is squarefree, and prime $\rm\:p\:|\:n\,\Rightarrow\, \color{#c00}{p\!-\!1\mid e\!-\!1}$ 
Proof $\ $ See this answer for a short simple proof.
A: $2730 = 2\cdot 3\cdot 5\cdot 7\cdot 13$
The Carmichael function or least universal exponent function is composed by least common multiple over prime components, so  $\lambda(2730) = \text{lcm}(\lambda(2), \lambda(3), \lambda(5), \lambda(7), \lambda(13)) = \text{lcm}(1,2,4,6,12)=12$. Note also that $2730$ is square-free, so the exponent cycle will be entered by the first power ($n^1$) for all numbers. Of course numbers coprime to $2730$ enter the cycle at the zeroth power ($1$).
Thus(!) $n^{(1+12)}\equiv n^1 \bmod 2730$ as required.
A: 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.
A: 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.
A: Here is some way of automating Rob John's method:
You can always use decomposition over polynomials :$\quad\displaystyle \Pi_k(n)=k!\binom nk=\prod\limits_{i=0}^{k-1} (n-i)$
e.g. $\ \Pi_4(n)=(n-3)(n-2)(n-1)n$

To solve problems of the type: "show $m$ divides the polynomial $P(n)$"

Let have a look at a simpler example using this technique :
Prove $(n^5-n)$ is divisible by 5 by induction.
Though for $n^{13}-n$ it is a bit tedious.
Here is a maple procedure to do it:
> binomexpansion :=proc(P)
local a,b,c,d,i,p,q:
p:=P: d:= degree(P):
printf("%a = ",P):
for i from d to 0 by -1 do
a:=coeff(p,n,i):
q:=expand(binomial(n,i)):
b:=coeff(q,n,i):
c:=a/b:
p:=p-c*q:
if((c>0)and(i<d)) then printf("+"); fi:
if(c<>0) then printf("%d(n,%d)",c,i); fi:
end do: printf("\n"):
end proc

> binomexpansion(n^13-n);

n^13-n = 6227020800(n,13)+37362124800(n,12)+97037740800(n,11)+142702560000(n,10)+
         130456085760(n,9)+76592355840(n,8)+28805736960(n,7)+6711344640(n,6)+
         901020120(n,5)+60780720(n,4)+1569750(n,3)+8190(n,2)

You can notice that all coefficients are divisible by $2730=2.3.5.7.13$.
And since the binomial coefficients are integers, you get the divisibility by $2730$ as a result.
