$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1  $ is divisible by  $ \ \large 2^{n+2} $ .
Answer:
For  $ n=1 \ $ we have 
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1. 
For $ n=2 $ we have 
$ \large 3^{2^2}-1=81-1=80  \ \ and \ \ 2^{2+2}=16 \ $
$Also \ \ \  16 /80 $ . 
Thus the statement hold for $ n=2 $ also. 
Let the statement hold for $ n=m \ $
Then $ a_m=3^{2^m}-1 \ \ is \ \ divisible \ \ by \ \ b_m=2^{m+2} \ $ 
We have to show that $ b_{m+1}=2^{m+3}  \ $ divide $ \ \large a_{m+1}=3^{2^{m+1}}-1 \ $ 
But right here I am unable to solve . If there any help doing this ? 
Else any other method is applicable also.

The Carmichael function $\lambda(2^{n+2})=2^n$ for positive $n$. Thus for any odd number $k$, $k^{2^n}\equiv 1 \bmod  2^{n+2}$ and  thus $2^{n+2}$ divides $k^{2^n}- 1$.
