How to prove $p=2^{2^s}+1$ for some $s \in \Bbb N$

Question

If $p-1$ is a power of $2$ where $p$ is prime how to prove that $p=2^{2^s}+1$ for some $s \in \Bbb N$.

Surely, $p=2^k+1$ then what will we use next?

P.S: I got it finally. We have to use $(a-b)|(a^n -b^n)$ then we have to use that $(2^r+1)|(2^k+1)$ if $k=rs$ with $s$ odd; leading to a contradiction.

So I am using a tag of proof verification.

Answer 1

You should be more detailed in your argument. Since $p-1=2^k$ we have $p=2^k+1$. Assume now that $k=mn$ and continue like here:

If $2^n+1$ is prime, why must $n$ be a power of $2$?

So $p$ is not prime, if $k$ is not a power of $2$. Hence we have $k=2^s$ and $p=2^{2^s}+1$.

Answer 2

If $k$ contains an odd prime $p$; let $k=pd$ then note that

$(2^{d})^{p}-(-1)^{p}$ = $(2^{d}+1)(\sum_{j=0}^{j=p-1}{2^{dj}(-1)^{j}}$)