$k=2^n + 1$ is prime $\rightarrow n=2^m$ I am struggling with this proof. I want to prove the contrapositive, $n=2^ab \rightarrow 2^n + 1$ is composite.
My professor gave me a hint, 
$n=2^ab$, $b=2r+1 \ge 3$  $\rightarrow$  $2^{2^a}+1 | 2^n+1$.  I truly don't know what I am doing and need lots of explanation.

Recall that if $b$ is odd, then
$$(x+1) \vert x^{b}+1$$
This can be seen immediately from the remainder theorem, since $(-1)^b + 1 = -1 + 1 = 0$. Equivalently, for odd $b$, we have
$$x^b+1 = (x+1)(1-x+x^2-x^3 \pm \cdots - x^{b-2} + x^{b-1})$$
Hence, if we write $n = 2^a \cdot b$, such that $2^a \Vert n$ i.e. $b$ is odd, we have $x^n+1 = x^{2^a \cdot b} + 1 = \left(x^{2^a}\right)^b + 1$ is divisible by $x^{2^a}+1$. Hence, for $x^n+1$ to be a prime, we need $n = 2^a$ i.e. $n$ is a power of $2$.
