TITLE: Why are primitive roots of unity the only solution to these equations?
QUESTION [7 upvotes]: I was led by this question to the following problem:
Find $n$ complex numbers $\lambda_1\dots\lambda_n\in\mathbb{C}$ that satisfy
$$\begin{align} 
\sum_i\lambda_i & =0\\
\sum_i\lambda_i^2 & =0\\
\sum_i\lambda_i^3 & =0\\
&\vdots\\
\sum_i\lambda_i^{n-1} & =0\\
\end{align}$$
I'm pretty sure that the only solution for these equations is $\lambda_k = w \zeta^k$ where $w$ is some complex number and $\zeta$ is the primitive $n^{\text{th}}$ root of unity. By "only" I mean of course that you can permute the $\lambda_i$'s because the equations are all symmetric polynomials, but that's it. 
Is there a proof for this?

REPLY [9 votes]: By writing the  elementary symmetric functions in terms of your power sums using Newton's identities, you get $$(X-\lambda_1)\cdots(X-\lambda_n)=X^n+(-1)^n\lambda_1\cdots \lambda_n=X^n-w^n=(X-w)(X-w\zeta)\cdots(X-w\zeta^{n-1})$$ where $\zeta$ is a primitive $n$th root of unity and $w$ is any $n$th root of $-(-1)^{n}\lambda_1\cdots \lambda_n$. The result follows at once now.
This solution also follows from the observation that the power sums are homogeneous and hence one can normalize the problem by dividing all $\lambda_i$ by some factor. Choosing the factor such that $(-1)^{n}\lambda_1\cdots \lambda_n=-1$ gives $X^n-1$.
[Thanks to Greg Martin for simplifying my original answer.]