Prove identity related to nths root of unity If $1=z_0,z_1,...,z_{n-1}$ are nth roots of unity, prove that
$$(z-z_1)(z-z_2)...(z-z_{n-1})=1+z+z^2+...+z^{n-1}$$
I don't know what is meant by the condition given. If I substitute $1=z_0,z_1,...,z_{n-1}$, isn't it become
$$(z-1)^{n-1}=1+z+z^2+...+z^{n-1}$$
Does this holds?
 A: Consider the equation 
$$z^n-1=0$$
The roots of this equation are the $n^{th}$ roots of unity, which let's say are 1,$z_1$,$z_2$,...,$\, z_{n-1}$. Then the expression on the LHS can be rewritten as the product of its factors:
$$(z-1)(z-z_1)(z-z_2)...(z-z_{n-1})=z^n-1$$
Taking $(z-1)$ to RHS,
$$(z-z_1)(z-z_2)...(z-z_{n-1})={z^n-1 \over z-1}$$
Notice that the RHS has now become the sum of the geometric series $1+z+z^2+$...
Which proves the result
$$(z-z_1)(z-z_2)...(z-z_{n-1})=1+z+z^2+z^3+...+z^{n-1}$$
You cannot substitute 1 for all the $n$ roots. Only one root can be one and the rest are complex, excluding -1.
A: You know that your roots are the solutions of $z^n-1=0$, hence
$$z^n-1=(z-z_0)(z-z_1)(z-z_2)...(z-z_{n-1})=(z-1)(z-z_1)(z-z_2)...(z-z_{n-1})$$
Now you just have to divide by $z-1$
$$\frac{z^n-1}{z-1}=(z-z_1)(z-z_2)...(z-z_{n-1})$$
but the l.h.s. is the partial sum of the geometric series, i.e. $\sum_{k=0}^{n-1}z^k$, hence
$$\sum_{k=0}^{n-1}z^k=(z-z_1)(z-z_2)...(z-z_{n-1})\ .$$
If you don't know the result on the geometric series you can compute the ratio directly or try to proove it.
