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?

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.
