Shows that $3^{n+1}$ divides $2^{3^{n}} + 1$ for each integer $n \geq 0$.
I have the inductive base which is prove that it is true for
$ n \in \mathbb{N}$
$3^{(1)+1}$ divides $2^{3^{(1)}} +1 $
$\frac{9}{9}$ is true
now I proposed that it is true for $n=n+1$
$3^{(n+1)+1}$ divides $2^{3^{(n+1)}} + 1$
how can i continue with the demonstration, help please
