proof factoring of $x^n-1$ I have a question about the proof of the theorem stating that $$x^n-1 = \prod_{i \in I} m^{(i)}(x)$$ is a factorisation in irreducible factors of $x^n-1$ over a field $\mathbb{F}_q$. Here $m^{(i)}$ is the minimal polynomial of $\beta^i$ where $\beta$ is a primitive $n^{th}$ unity root (I don't know whether this is the right expression in English) and $I \subseteq \{0, 1, \ldots, n-1\}$ so that it contains exactly 1 element from every cyclotomic coset.
Anyway, my question is about a certain point in the proof where there's standing $$\prod_{k =0}^{n-1} (x-\beta^k) = x^n-1.$$
Why should I believe this is true for an extension-field $\mathbb{F}$ of $\mathbb{F}_q$?

By definition, if $\beta$ is a primitive $n$-th root of unity, then all $n$-th roots of unity are given by $\beta^k$, for $k=0,\dots,n-1$. The equation
$\prod_{k =0}^{n-1} (x-\beta^k) = x^n-1$ is just this statement, since an $n$-th root of unity is a root of the polynomial $x^n-1$. The equation expresses a polynomial factorized in terms of its roots.
