I need some help please. This is for a coding theory course dealing with Galois Fields $GF(2^r)$, the problem is stated as follows:
Show that if $h(x) \in \mathbb{F}_2[x]$ is an irreducible polynomial of degree $n$, then $h(x)$ divides $1+x^m$ for some $m \leq 2^m-1$.
So far I have $0 \leq m<2^n-1$, there are $2^n-1$ possible non-zero remainders for
$$x^m+1 = a(x)h(x)+r(x)$$
However, when $r(x)=0$, how does that specifically imply that $m=2^n-1$? Is there another route to follow here? How can I show that $x^{2^n-1} \equiv 1 \mod(h(x))$?