Prove that the $GCD(2^{k}, n) = 1$ given that $GCD(2, n) = 1 $ and n is odd Not too sure how to approach this problem, 

If  $GCD(2, n) = 1 $, this implies that $2x + ny = 1$ for some integer $x,y$. How do we show the statement given this information? (If it is in fact possible)

Thanks
Hint: I wouldn't use Bezout's here (although you probably can). Assume $\gcd(2^k,n)\neq 1$. Then there exists a prime $p$ that divides both $2^k$ and $n$. What options are there for that prime?
If ( 2 and n ) have  greatest common divisor only 1 ,then obviously  ( 2*2*2.... ) and n will have greatest common divisor 1 So , gcd ( 2 ^ k , n) =1