The $m$-th Fermat number is defined by $F_m = 2^{2^m}+1$ (where $m\geq 0$ is an integer).
The condition that we must show is equivalent to $2^{F_n} = 2^{2^{2^n}+1} \equiv 2$ modulo $F_m$. This also shows that the order of $2$ modulo $F_m$ divides $F_n = 2^{2^n}+1$. However, what I find weird or unintuitive is how to use the inequality $m \leq 2^n-1$. Usually, when showing divisibilities, some inequalites turn out to be difficult obstables.
I believe this property of Fermat numbers may be helpful: $$ F_m = \left( \prod_{i=0}^{m-1} F_i \right) + 2 $$ where $m \geq 1$ because it was also on the same exercise sheet. Still, I cannot relate this to the inequality assumption too.
Do you have any idea or helpful remarks to guide me through this problem? Any help is appreciated!