$\sum_i x_i^n = 0$ for all $n$ implies $x_i = 0$ Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost.
Let $x_1 , x_2, \dots, x_k$ be complex numbers satisfying:
$$x_1 + x_2+ \dots + x_k = 0$$
$$x_1^2 + x_2^2+ \dots + x_k^2 = 0$$
$$x_1^3 + x_2^3+ \dots + x_k^3 = 0$$
$$\dots$$
Then $x_1 = x_2 = \dots = x_k = 0$.

The statement seems obvious because we have more than $k$ constraints (constraints that are in some sense, "independent") on $k$ variables, so they should determine the variables uniquely.  But my attempts so far of formalizing this intuition have failed.  So, how do you prove this statement?  Is there a generalization of my intuition?
 A: Apply Newton's identities. This shows the power sums form a basis for the symmetric polynomials, since the elementary symmetric polynomials are a basis and the elementary symmetric polynomials can be represented as linear combinations of power sums. So all symmetric polynomials evaluate to $0$ at the $x_i$. Consider a polynomial with the $x_i$ as roots. The coefficients are the elementary symmetry polynomials in the $x_i$, so they are all zero. The polynomial is then $x^k$, which shows $x_i=0$ for all $i$.
A: For a slightly different method than Potato's second answer (but the idea is mainly the same):
Without loss of generality, the system of equations can be written as: $$\left\{ \begin{array}{lcl} \lambda_1x_1 + \lambda_2x_2+ \dots + \lambda_k x_k &= &0 \\ \lambda_1x_1^2 + \lambda_2x_2^2+ \dots + \lambda_k x_k^2 & = & 0 \\ & \vdots & \\ \lambda_1x_1^k + \lambda_2x_2^k+ \dots + \lambda_k x_k^k & = & 0 \end{array} \right.$$
where $\lambda_i>0$, $x_i \neq 0$ and $x_i \neq x_j$ for $i \neq j$. Indeed, if $x_i=x_j$ replace $x_i+x_j$ with $2x_i$ and if $x_i=0$ just remove it. By contradiction, suppose $k \geq 1$.
Now, the family $\{ (\lambda_1 x_1^j, \dots , \lambda_k x_k^j) \mid 1 \leq j \leq k \}$ cannot be linearly independent since the vector space $\{(y_1,\dots, y_k) \mid y_1+ \dots+ y_k=0 \}$ has dimension $k-1$ (it is a hyperplane). Therefore, the matrix
$$A:=\left( \begin{matrix} \lambda_1x_1 &  \lambda_2x_2 & \dots & \lambda_k x_k  \\ \lambda_1x_1^2 & \lambda_2x_2^2 & \dots & \lambda_k x_k^2  \\ \vdots & \vdots & & \vdots  \\ \lambda_1x_1^k & \lambda_2x_2^k & \dots & \lambda_k x_k^k  \end{matrix} \right)$$
is not invertible. Using Vandermonde formula, $$0= \det(A)=  \prod\limits_{i=1}^k \lambda_i   \cdot \prod\limits_{i=1}^k x_i  \cdot \prod\limits_{i<j} (x_i-x_j) $$
which is nonzero by assumption.
A: Rename the $x_i$ as $\alpha_i$ so that I can borrow the notation of this wikipedia article. Multiply the first row of the matrix in the article by $\alpha_1$, the second by $\alpha_2$, and so on. This gives a matrix with determinant
$$\left( \prod_i \alpha_i \right)\left(\prod_{i< j} (\alpha_i-\alpha_j) \right).$$
Since the homogeneous equation with these coefficients has a solution (the vector $(1,1,\cdots,1)$), the determinant must be zero. So either $\alpha_i=0$ for some $i$ or $\alpha_i=\alpha_j$ for some pair of distinct indices. In the first case, we reduce to the same problem with $k-1$ terms and can induct. In the second case, we can consider a slightly modified problem: identifying the $i$th and $j$th term, we have the same problem with $n-1$ variables, except the coefficient of $\alpha_j$ is $2$. But, since this still gives a nonzero solution to the corresponding homogeneous equation, we can again "induct" using the method described above. The number of variables must eventually decrease to one, from which we see $\alpha_1=0$, and hence $x_i=0$ for all $i$ in the original problem. 
