Can the expression be simplified?

Question

If $1, a_1, a_2,\ldots, a_{n-1}$ are $n$-th roots of unity, can the following expression be simplified?

$(1-a_1)(1-a_2)\cdots(1-a_{n-1})$?

Answer 1

We know that

$$x^{n-1}+x^{n-2}+\cdots+x+1=(x-a_{1})(x-a_{2})\cdots(x-a_{n-1}).$$

That's because $a_{1},a_{2},\cdots,a_{n-1}$ are the distinct roots of the equation $$x^{n-1}+x^{n-2}+\cdots+x+1=0$$

Put $x=1$ and get

$$(1-a_{1})(1-a_{2})\cdots(1-a_{n-1})=1+1+\cdots+1=n$$

Answer 2

If $a_0=1,a_1,...,a_{n-1}$ are all the distinct $n$-th roots of unity, then $$z^n-1=(z-1)(z-a_1)\cdot...\cdot (z-a_{n-1})$$ Divide both sides by $z-1$, and take a limit when $z\to1$. What do you get?