$a_i$ are the n-th roots of $1\in\mathbb{C}$, why does $(1-a_2)\cdot...\cdot(1-a_n)=n$?

Question

For $1<i\leq n$, let $a_i$ be the n-th roots of $1\in\mathbb{C}$, why does $(1-a_2)\cdot...\cdot(1-a_n)=n$?

Answer 1

$$z^n - 1 = (z-1)(z^{n-1} + \cdots + z + 1) = (z-1)\prod_{i=2}^n(z-a_i)$$

$$\Rightarrow z^{n-1} + \cdots + z + 1 = \prod_{i=2}^n(z-a_i) \stackrel{z=1}{\Rightarrow} n = \prod_{i=2}^n(1-a_i)$$

Answer 2

L'hopital's (sp?) rule applied to $\lim_{x \to 1} \dfrac{x^n-1}{x-1}$.