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: 
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

$$(1-a\_{1})(1-a\_{2})\cdots(1-a\_{n-1})=1+1+\cdots+1=n$$
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?