Originally Posted by
NonCommAlg the claim would be true if we also assume that n > 1. the proof is quite easy: let $\displaystyle f_n(k)=k^n + k^{n-1} + \cdots + k + 1, \ \ k > 1, \ n>1.$ it's clear that if $\displaystyle n$ is even, then $\displaystyle f_n(k)$ is odd.
so we may assume that $\displaystyle n$ is odd. the proof now is by induction over $\displaystyle n.$ we have $\displaystyle f_3(k)=k^3+k^2+k+1=(k+1)(k^2+1).$ suppose $\displaystyle f_3(k)$ is a power of 2. then $\displaystyle k=2r+1,$ for
some $\displaystyle r \geq 1,$ and so $\displaystyle k^2+1=4r(r+1)+2,$ which is never a power of 2 because $\displaystyle r > 0.$ if $\displaystyle n=2m+1 > 3,$ then the identity $\displaystyle f_{2m+1}(k)=f_m(k) (k^{m+1} + 1)$ completes the proof.