If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$, prove that $(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$

Question

My first post here

So I was doing some past year questions and this one popped out and I haven't been able to progress much upon it and I believe there's a trick which will get it done in no time.

If the roots of the equation
$x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$

we need to show that $$(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$$

Can I be helped with some hints to push me in the right direction to solve it?

Thanks in advance.

Answer 1

Apparently

$ x^n-1 = (x-1)(x-\alpha)(x-\beta)(x-\gamma)... $

setting $P(x) = \tfrac{x^n-1}{x-1}$

$ P(x)= (x-\alpha)(x-\beta)(x-\gamma)... $

Now just evaluate at x = 1.