If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$, prove that $(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$ 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.
 A: 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.
A: Denote the roots by $x_0 = 1, x_1, x_2, x_3, \dots, x_{n-1}$
(roots of unity, usual indexing in counter-clockwise direction)
Then $x_k = x_1^k$
So this whole product is an expression involving just $x_1$
And from the roots of unity theory we know that
$$x_1 = \cos\frac{2\pi}{n} + i \cdot \sin\frac{2\pi}{n}$$
Proceeding from here is a matter of doing some simple algebra.
A: If the roots are $r_i$, $i=1\ldots n$, with $r_1=1$, we have
$\prod\limits_{i=1}^{n}r_i= -(-1)^n$
$\sum\limits_{i=1}^{n}r_i = \sum\limits_{i<j}^{n}r_ir_j = \sum\limits_{i<j<k}^{n}r_ir_jr_k =\ldots= 0$ (till taking $n-1$ terms at a time)
Now, $\sum\limits_{i=2}^{n}r_i=\sum\limits_{i=1}^{n}r_i-1=0-1=-1$
Also, $\sum\limits_{i,j=2,i<j}^{n}r_ir_j=\sum\limits_{i,j=1,i<j}^{n}r_ir_j-\sum\limits_{i=2}^{n}r_i=0-(-1)=1$
We can show that the terms $\sum\limits_{i=2}^{n}r_i, \sum\limits_{i,j=2,i<j}^{n}r_ir_j,\sum\limits_{i,j,k=2,i<j<k}^{n}r_ir_jr_k, \ldots$
have alternating values $-1$ and $1$
$\therefore (1-\alpha)(1-\beta)(1-\gamma)\ldots$
$=\prod\limits_{i=2}^{n}(1-r_i)$
$=1-\sum\limits_{i=2}^{n}r_i+\sum\limits_{i,j=2}^{n}r_ir_j-\ldots$
$=1-(-1)+1-(-1)+\ldots=n$
A: Like Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity OR Problem based on sum of reciprocal of $n^{th}$ roots of unity
using Transformation of equations, let $1-x=y$
$$\implies(1-y)^n=1\iff  y^n-\binom n1y^{n-1}+\cdots+\binom n{n-1}(-1)^{n-1}y=0$$
whose roots are $y_k, 0\le k\le n-1$
Clearly, one of the them is $0,$ let $y_0=0$
So, the roots of $$y^{n-1}-\binom n1y^{n-2}+\cdots+\binom n{n-1}(-1)^{n-1}=0$$ will be $y_k, 1\le k\le n-1$
Finally use Vieta's formula
