Here's the full problem:
Let $\mathbb{F}_{p}$ denote the finite field of size $p$, where $p$ an integer prime greater than $2$. Suppose that $f(x)=x^{m}+1$ is an irreducible polynomial in $\mathbb{F}_{p}[x]$ and let $E$ be its splitting field over $\mathbb{F}_{p}$.
a) Show that for every root $\alpha$ of $f(x)$ in $E$ we have $\alpha^{2m}=1$ and $\alpha^{k}\neq 1$ for any $1\leq k <2m$.
b) Prove that $2m$ divides $p^{m}-1$, but that $2m$ does not divide $p^{n}-1$ for any $1\leq n <m$. (Hint: consider the size and nature of the multiplicative group of $\mathbb{F}_{p^m}$ and use Lagrange's theorem).
So I am done with part (a) and am stuck on part (b). My question is, how does part (b) follow from part (a)? I'm not sure how to bring $\mathbb{F}_{p^m}$ into part (b).
Let $\alpha$ be a root of $f$. Then $[\mathbb F_p(\alpha):\mathbb F_p]=m$, so $\mathbb F_p(\alpha)=\mathbb F_{p^m}$. This is also the splitting field of $f$ since $f(\alpha)=0$ implies $f(\alpha^{p^i})=0$ for all $0\le i<m$.
From a) you know that the order of $\alpha$ in the multiplicative group $\mathbb F_{p^m}^\times$ is $2m$, so $2m\mid p^m-1$ (by Lagrange). Now suppose $2m\mid p^n-1$ with $n<m$. Since $\alpha^{2m}=1$ we have $\alpha^{p^n-1}=1$, so $\alpha^{p^n}=\alpha$. It follows that every $x\in\mathbb F_{p^m}$ satisfies $x^{p^n}=x$, so $\mathbb F_{p^m}=\mathbb F_{p^n}$, a contradiction.