If $1,a_1,a_2,...,a_{n-1}$ are the $n$ roots of $1$,then $(1-a_1)(1-a_2)...(1-a_{n-1}) \space ?$

Question

If $1,a_1,a_2,...,a_{n-1}$ are the $n$ roots of unity, then how can we find the value of $$(1-a_1)(1-a_2)...(1-a_{n-1}) \space ?$$

My Approach:
If $n=2$ , then $1,-1$ are the roots of unity
$\therefore (1-a_1)=(1-(-1))=2$
for $n=3 \space :$ $1,\omega,\omega^2$ are the roots of unity
$\therefore (1-a_1)(1-a_2)=(1-\omega)(1-\omega^2)$
$\quad \quad \quad =1-\omega^2 -\omega +1=3$
so we conclude for $n$ the value $(1-a_1)(1-a_2)...(1-a_{n-1}) \space = n$

but i want a direct process to evaluate $(1-a_1)(1-a_2)...(1-a_{n-1}) $ (without generalisation),so how can i do so?

Answer 1

Consider the polynomial $x^n-1=(x-1)(x-a_1)...(x-a_{n-1})$. Divide by $x-1$ to get the polynomial $x^{n-1}+x^{n-2}+...+x+1$. Evaluate at $x=1$ to get the answer $n$.