Let $q$ be a prime power and $n$ a positive integer s.t. $\gcd(n, q) = 1$. Let $E(n)$ be the set of the complex $n$-th roots of unity. For every positive integer $d$ such that $d | n$ , let
$$Q_d(x) := \prod_{\alpha \in E(n):\mathrm{ord}(\alpha)=d} (x- \alpha)$$
Why does it hold true that $Q_n(x) = \prod_{d:d|n} (x^d -1 )^{\mu(n/d)} = \prod_{d:d|n}(x^{n/d}-1)^{\mu(d)}$, where
$\mu: \mathbb Z_{>0} \to \{-1, 0, 1\}$ is the Möbius function.
Thanks for any help!
First, note that $$ \prod_{d\mid n} Q_d(x) = \prod_{\alpha\in E(n)} (x - \alpha) = x^n - 1. $$ The result is then an application of the Möbius inversion formula written in multiplicative form, namely $$ F(n) = \prod_{d\mid n} f(n) \quad \Rightarrow \quad f(n) = \prod_{d\mid n} F(n/d)^{\mu(d)}, $$ where we take $f(n) = Q_n(x)$ and $F(n) = x^n - 1$.
Incidentally, $Q_n(x)$ is more commonly written $\Phi_n(x)$ and is called the $n^\text{th}$ cyclotomic polynomial.