nth roots of unity - meaning of polynomial equation? I am reading above the roots of unity, given by
$$
w^n = 1 \Rightarrow w = cis(\frac{2k\pi}{n})
$$
Now let $k = 1$. I read that because w satisfies $ w^n = 1$ and thus (for some reason I do not understand), $w^n - 1 = 0$ has a factor $w - 1$ and dividing $w-1$ into $w^n - 1$ gives a factorization
$$
(w-1)(w^{n-1}+w^{n-2}+\cdot\cdot\cdot + w + 1) = 0
$$
Then since $w \neq 1$, $(w^{n-1}+w^{n-2}+\cdot\cdot\cdot + w + 1) = 0$. 
So finally to my questions:


*

*Why is $w \neq 1$? I know that $w$ raised to the power of n is 1, but $w$ could still be 1, I think? Especially with $w = cis(\frac{2\pi}{1}) = cos(\frac{2\pi}{1}) + sin(\frac{2\pi}{1}) = 1 + 0 = 1$. If so, $w = 1$? 

*What can I use this for? What does it show? That $(w^{n-1}+w^{n-2}+\cdot\cdot\cdot + w + 1) = 0$? I know that the length of all the roots is $1$ and thus all are located on the unit circle - but is this related to the polynomial equation? If so, how?


Part of the book that I am reading:

 A: OP is looking for a geometrical meaning of $z^{n-1} + z^{n-2} + ... + z + 1 = 0$, which is known to be true for $n$th roots of unity other than $1$. One way to interpret this is as follows. Since, as OP pointed out, each $z^{k}$ not only has unit length but is a $n$th root of unity in itself, the equation is the sum of $n$ different $n$th roots of unity. The equation then implies that their sum is always $0$. Is this fact alone surprising? I would say not necessarily. However, it might become more interesting when one starts to look at the properties of the numbers ${z^{n-1}, z^{n-2}, ..., z, 1}$ as a function of $n$. For example, if $n$ is even, they are symmetric with respect to the origin so one can always rearrange the equation so that every two terms add up two zero. If $n$ is odd, there is no such arrangement and the fact that they all add up to $0$ together is a bit more interesting.
In the end, given without context, this fact will only be interesting in so far as it is used to elucidate other properties of roots of unity (and the myriad applications they have across mathematics). Hope this helps.
A: For any n, there exist n values of z, such that  $z^n= 1$, one of which is z= 1.  You don't give the full statement.  Where it says "$z\ne 1$", the text must have said they were looking for other roots of 1.
As for what it tells you- it tells you that the nth roots of 1, other than 1 itself, satisfy the equation $z^{n-1}+ z^{n-2}+ \cdot\cdot\cdot+ z+ 1= 0$.
By the way, yes, the roots (including z= 1) lie on the unit circle and they are equally spaced along that circle.
A: $$(z-1)(z^{n-1}+z^{n-1}+\cdots 1)=z^n-1$$
is a general polynomial factorization and is true for any $z$.
From this, we deduce that $\omega_0=1$ is certainly a solution of
$$\omega^n-1=0.$$
And if we consider another solution, i.e. $\omega_k\ne1$, we can affirm that any root of 
$$\omega^n-1=0$$ is also a root of 
$$\omega^{n-1}+\omega^{n-2}+\cdots 1=0.$$
(By the way, for $\omega_0=1$, the LHS equals $n$.)
