How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$ where $n\in\mathbb{N}-\{0\}$ with induction? So the question is as above is shown:

How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$?

$n=1: 8^2 - 5^2 = 39$ is a multiple of $13$.
I.H.: Suppose $8^{2^n} - 5^{2^n}$ is divisible by $13$ is true $\forall n \in \mathbb{N}\setminus\{0\}$. 
I got stuck on this $n= m + 1$ case.
$8^{2^{m+1}} - 5^{2^{m+1}}= 8^{2^m\cdot 2}- 5^{2^m\cdot 2}$
Can somebody please help me?

More generally, $8^{m} - 5^{m}$ is divisible by $13$ when $m$ is even.
Indeed, $8^{m} - 5^{m} = (13-5)^m-5^m = 13a+5^m-5^m=13a$ by the binomial theorem.
Here is a proof by induction:
$8^{m} - 5^{m}=13a$
$8^{2} - 5^{2}=13b$
$8^{m+2} - 5^{m+2} = 8^m 8^2 - 5^m 5^2 = (13a+5^m)8^2 - 5^m 5^2 = 13c +  5^m  8^2 - 5^m 5^2 = 13c +5^m(8^2 - 5^2) = 13c + 5^m 13b = 13d$ 
