n^13-n=n(n^12-1).

We will rely on Fermat's Little Theorem.

If p|n there is nothing to prove.

We will assume that p doth not divide n.

That is, prove 2,3,5,7,13 divide n^12-1

First, the case p=2 is trivial.

If p=3 then,

n^2=1(mod 3)

Raise both to the power of 6,

n^12=1(mod 3)

If p=5 then,

n^4=1(mod 3)

Raise both to the power of 3,

n^12=1(mod 3)

If p=7 then,

n^6=1(mod 7)

Raise both to the power of 2,

n^12=1(mod 7)

If p=13 then,

n^12=1(mod 13)

Q.E.D.