Factorization of $x^{rn} - 1$

Question

Let $q$ be a prime power and $r$ a divisor of $q-1$. Let $n = \frac{q^m-1}{r}$ and $N = rn = q^m-1$. Denote by $\beta$ a primitive element of $\mathrm{GF}(q^m)$ and $\lambda = \beta^{n}$, then $\lambda$ is an element of $\mathrm{GF}(q)^*$ with order $r$. The question is why $$ x^{rn} - 1 = \prod\limits_{i=0}^{r-1}(x^n - \lambda^i) $$ and $\gcd(x^n - \lambda^i, x^n - \lambda^j) = 1$ for $0 \le i \ne j \le r-1$.

Answer 1

The notation obscures the simple fact that the set of roots of the parenthesized expression on the right is exactly $M_i = \{\beta^i, \beta^{i+r}, \ldots, \beta^{i+(n-1) r} \}$ (just substitute them!) Then, for different $i, j \in \{0, 1, \ldots, r-1\}$, $i \not = j$, we have $M_i \cap M_j = \varnothing$, and this simultaneously proves the product equality and the statement about GCD.