Chapter 4.3, Problem 10E

Show that if ${\text{2}}^m\text{+1}$ is and odd prime, then $m={\text{2}}^n$ for some nonnegative integer n. [Hint: First show that the polynomial identity $x^m+1=(x^k+1)(x^k{}^{\text{(}t-1)}-x^{k(t-\text{2})}+\dots -x^k+\text{1})$ holds, where m = kt and t is odd.]