Proving divisibility of sequences 
For $a_n=6^{(2^n)}+1$, how can I show that $a_n\mid(a_{n+1}-2)$ ?


Observe that $$a_{n+1}-2=6^{2^{n+1}}+1-2$$ $$=6^{2^{n}\cdot2}-1$$ $$=\left(6^{2^{n}}\right)^2-1$$ $$=\left(6^{2^{n}}-1\right)\left(6^{2^{n}}+1\right)$$ $$=a_n\left(6^{2^{n}}-1\right)$$ So, $\left(a_{n+1}-2\right)$ is a multiple of $a_n$ 
$\therefore$ $$a_{n+1}-2=a_n\left({6^2}^n-1\right)$$
