Find GCD

We can express $n^{13}-n$ as $(n-1)n(n+1)(n^2-n+1)(n^2+1)(n^2+n+1)(n^4-n^2+1)$ . Of these, $(n-1)n(n+1)$ is always divisible by $3!=6$ , $(n-1)n(n+1)(n^2+1)$ is always divisible by $6\times 5=30$ , $(n-1)n(n+1)(n^2+1)(n^2-n+1)(n^2+n+1)$ is always divisible by $30\times 7=210$ , and the entire product is always divisible by $210\times {13}=2730$ . (Using the standard divisibility conditions). Therefore $n^{13}-n$ is divisible by $2730$ for all positive integral values of $n$ . Hence the required GCD is $\boxed {2730}$

Let $d$ be the g.c.d.

Now $d | {n^{13}-n}$ $\forall n =2,..,13$

$2^{13}-2 = 2 \times 5 \times 7 \times 9 \times 13$

$=> d | (2 \times 5 \times 7 \times 9 \times 13)$

Observe that $2,3,5,7,13$ all divide ${n^{13}-n}$ $\forall n =2,..,13$ { using Euler Totient function}

and $9$ doesn't divide ${3^{13}-3}$ as ${3^{12}-1}$ is not divisible by $3$ .

Hence $d = 2 \times 3 \times 5 \times 7 \times 13 = \boxed{2730}$ .