Irreducible polynomials over GF(2) of degree m divide this?

Question

I've tried so much to prove the following problem but it was waste of time. I wonder if anyone could help me.

Any irreducible polynomial over $GF(2)$ of degree $m$ divides $X^{2^m−1}\ +\ \ 1$

Answer 1

If $f$ is an irreducible polynomial of degree $m$, then $F=GF(2)[X]/(f)$ is a field with $2^m$ elements. Let $F=GF(2)(\alpha)$. Apply Lagrange's theorem to the multiplicative group $F^\times$ to conclude that $\alpha^{2^m-1}=1$. Since $f$ is the minimal polynomial of $\alpha$, the result follows. (Note that $-1=1$ in $F$.)

Indeed, all polynomials $g$ such that $g(\alpha)=0$ are multiples of $f$. In other words, $f$ generates the ideal of polynomials that have $\alpha$ as a zero.