Product of Differences between 1 and Complex Roots of Unity

Theorem
Let $\alpha$ be a primitive complex $n$th root of unity.

Then:
 * $\ds \prod_{k \mathop = 1}^{n - 1} \paren {1 - \alpha^k} = n$

Proof
From Power of Complex Number minus 1: Corollary:
 * $\ds \sum_{k \mathop = 0}^{n - 1} z^k = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}$

The result follows by setting $z = 1$.