Prove that $2^{n+2}$ is a divisor of the number $k^{2^{n}}-1$ Question : 
If n is a natural number and k is an odd number 
$k\in \mathbb{Z}$ then show that $2^{n+2}$ is a divisor of
the number $k^{2^{n}}-1$
My try as following : 
I'm think use induction : 
$n=1$ then $2^{3}=8$ since $k$ is odd so : $k^{2}=1\mod 8$ 
We find : $2^{3}$ divisor of $k^{2}-1$ (correct )
Now for $n+1$ how I prove $2^{n+3}\mid (k^{2^{n+1}}-1)$ 
And if can prove it without induction ??

The trick is $k^{2*m} - 1 = (k^m -1)(k^m+1)$ and if $2^z\mid k^m -1$ then $k^m$ is odd and $k^m + 1$ is even so $2^z|k^m-1$ and $2\mid k^m+1$ so $2^z*2\mid (k^m-1)(k^m+1)$.
So by induction:
Base case: If $k$ is odd the $k = 2m+1$ for some $m$ and $(2m+1)^2-1= 4m^2 + 4m = 4m(m-1)$ either $m$ or $m-1$ is even so $2|m(m-1)$ and $2^3 = 8 = 4*2\mid 4*m(m-1)$ 
Induction case: So if ${2^{k-1}}\mid (k^{2^{k+1}}-1)$ then $k$ is odd and  $2^{2^k}=2^{2^{k-1}}*2\mid (k^{2^{k+1}}-1)(k^{2^{k+1}} + 1) = (k^{2^{k+2}} - 1)$.
