The solutions of a system of polynomials Given positive integers $m_1,...,m_n$, is it possible to solve the following equation system over the field of complex numbers?
$$m_1x_1+\cdots+m_nx_n=0$$
$$m_1x_1^2+\cdots+m_nx_n^2=0$$
$$\cdots$$
$$m_1x_1^{n-1}+\cdots+m_nx_n^{n-1}=0$$
$$x_1x_2\cdots x_n=1.$$
When $n=1,2,3,4$, I can find a formula for the solutions. But I can not do it for $n>4$. Also, when $m_1=m_2=...=m_n=1$, the solutions can be written as
$$(\zeta^{\sigma(1)},\cdots,\zeta^{\sigma(n)})$$
where $\zeta$ is a $n$-th primitive root of unity and $\sigma$ is an element of the symmetric group $S_n$.
For other concrete examples, the Mathematica numerical computation shows that the number of solutions to this equation system would be $n!$. Does anybody have an idea to prove it?
In general, if we do not assume $m_1,...,m_n$ are positive integers, what is the condition on $m_1,...,m_n$ such that this equation system has a solution?
 A: This is not a complete solution but it seems to me that it can be a good starting point. I will prove that the system has $n!$ distinct solutions for a general choice of the coefficients $m_i$. 
All your equations are homogeneous in the $x_i$, except the last one. Homogenize adding a variable $x_0$; then all equations remain unchanged except for the last one, which becomes $$x_1x_2 \cdots x_n-x_0^n=0.$$
Now you can see the solutions of your system as the intersections of $n$ hypersurfaces in $\mathbb{P}^n(\mathbb{C})$, with homogeneous coordinates $[x_0: \ldots :x_n]$. If these hypersurfaces intersect transversally, then Bézout Theorem implies that there are only a finite number of solutions in the projective space, and this number is the product of the degrees of the equations, namely $1\cdot 2\cdot 3 \cdots \ n= n!$.
On the other hand, for $m_1=m_2 = \ldots =m_n=1$ you find exactly $n!$ solutions. So you can conclude that for a general choice of the coefficients $(m_1, \ldots, m_n) \in \mathbb{C}^n$ your system has exactly $n!$ distinct solutions in $\mathbb{P}^n(\mathbb{C})$. 
Finally, these solution will be solutions of the original system if and only if all of them are outside the hyperplane at infinity $x_0=0$. This means that for any solution all the coordinates $x_1,\ldots, x_n$ must be nonzero. Again, you know that this happens in the case $m_1=m_2 = \ldots =m_n=1.$
So we can conclude that your system has exactly $n!$ solutions for a general choice of the coefficients $(m_1, \ldots, m_n) \in \mathbb{C}^n$, where general means that the coefficients can be chosen outside a Zariski closed subset $Z$ of $\mathbb{C}^n$. 
In particular the general $(m_1, \ldots, m_n) \in \mathbb{N}^n$ does not belong to $Z$, hence for a general choice of the positive integers $(m_1, \ldots, m_n)$ your system has precisely $n!$ distinct solutions.  
A: I believe this follows using Vandermonde matrices.  First I will prove a proposition that gives a bit more than what the OP asks.
For every integer $n \geq 2$, for every integer $r=1,\dots,n$, define
$$ F_r(X_1,\dots,X_n) = F_{n,r}(m_1,\dots,m_n;X_1,\dots,X_n) = m_1X_1^r + \dots + m_nX_n^r.$$  Let $P\subset \mathbb{C}$ be a nonempty subset that is stable under addition and which does not contain $0$, e.g., the set of positive integers.
Proposition $Q_n$.  For every integer $n\geq2$, for every choice of $m_1,\dots,m_n$ in $P$, every nonzero solution $(A_1,\dots,A_n)\neq (0,\dots,0)$ of $$
F_{n,1}(m_1,\dots,m_n;X_1,\dots,X_n)=\dots=F_{n,n-1}(m_1,\dots,m_n;X_1,\dots,X_n)=0$$ is not a solution of $F_{n,n}(m_1,\dots,m_n;X_1,\dots,X_n) = 0$, it has no coordinate equal to $0$, and it has no repeated coordinates.
Proof  This will be proved by induction on $n$.  It is straightforward to prove $Q_2$, the base case of the induction.  Basically it boils down to the fact that none of $m_1$, $m_2$ nor $m_1+m_2$ can equal $0$.  
Now by way of induction, assume that $n>2$ and that $Q_m$ holds for all integers $2\leq m < n$.  Let $(A_1,\dots,A_n)$ be a nonzero solution of the system $F_1=\dots=F_{n-1}=0$.  If any $A_i$ is zero, then after permuting the coordinates and $(m_1,\dots,m_n)$, we may assume that $A_n=0$. 
Then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system
$$ F_{n-1,1}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,\dots ,$$
$$ F_{n-1,n-2}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,$$
and which also solves the equation $F_{n-1,n-1}=0$.  This contradicts the induction hypothesis.  Thus, no $A_i$ equals $0$.  
Similarly, if any coordinates are repeated, say $A_{n-1}=A_n$, then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system,
$$ F_{n-1,1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0, \dots, $$
$$ F_{n-1,n-2}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0,$$
and which also solves the equation $$F_{n-1,n-1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1})=0.$$  Again this contradicts the induction hypothesis.  Thus, there are no repeated coordinates.
Therefore, let $(A_1,\dots,A_n)$ be an $n$-tuple with no zero coordinates and no repeated coordinates.  Then, by the theory of the Vandermonde determinant, there is only the trivial solution $(Y_1,\dots,Y_n)$ of the following linear system,
$$
\left\{ \begin{array}{rrrrrrr}
Y_1A_1   & + & \dots  & + & Y_nA_n   & = & 0, \\
Y_1A_1^2 & + & \dots  & + & Y_nA_n^2 & = & 0, \\
\vdots   &   & \ddots &   & \vdots   &   & \vdots \\
Y_1A_i^r & + & \dots  & + & Y_nA_n^r & = & 0, \\
\vdots   &   & \ddots &   & \vdots   &   & \vdots \\
Y_1A_1^n & + & \dots  & + & Y_nA_n^n & = & 0. 
\end{array} \right.$$  Indeed, the determinant of the coefficient matrix is
$$
\text{det}([A_s^r]_{1\leq r,s \leq n}) = \pm A_1\cdots A_n \prod_{k<l}(A_l-A_k),
$$
which is nonzero since all $A_i$ are nonzero and distinct.  Therefore, if $(m_1,\dots,m_n)$ is a solution, then all $m_i$ equal zero.  But this contradicts the hypothesis that all $m_i$ lie in $P$, which does not contain $0$.  Therefore, for $(A_1,\dots,A_n)$ as above, there is no choice of $(m_1,\dots,m_n)$ in $P$ such that all $F_{n,1} = \dots = F_{n,n-1}=F_{n,n}=0$.  Therefore, if $(A_1,\dots,A_n)$ is as above and solves $F_{n,1}=\dots=F_{n,n-1}=0$, then it does not solve $F_{n,n}=0$.  Therefore Property $Q_n$ holds.  Therefore, for every integer $n\geq 2$, Property $Q_n$ holds by induction on $n$.
End Proof of Proposition $Q_n$.
Combinining the proposition with Francesco's homogenization argument, there are $n!$ solutions of the original system, when counted with multiplicity.  The point is that, for Francesco's homogeneous system, $x_0$ cannot equal $0$, or that would force at least one of the coordinates $A_i$ to be zero, contradicting the proposition.
But, in fact, the multiplicity is always $1$.  If you form the Jacobian matrix of Francesco's system evaluated at a solution $(A_0,A_1,\dots,A_n)$, it is the following $n\times(n+1)$-matrix,
$$
\left[
\begin{array}{rr}
-nA_0^{n-1} & *   & * & \dots & * \\
0          & m_1 & m_2 & \dots & m_n \\
0          & 2m_1A_1 & 2m_2A_2 & \dots & 2m_n A_n \\
\vdots      & \vdots & \vdots & \ddots & \vdots \\
0          & rm_1A_1^{r-1} & rm_2A_2^{r-1} & \dots & rm_nA_n^{r-1} \\
\vdots     & \vdots        & \vdots        & \ddots & \vdots \\
0          & (n-1)m_1A_1^{n-2} & (n-1)m_2A_2^{n-2} & \dots & (n-1)m_nA_n^{n-2}
\end{array} \right].
$$
If you delete any one of the final $n$ columns, then the determinant of the remaining square matrix is,
$$
\pm n!m_1\cdots m_n A_0^{n-1}\prod_{k<l} (A'_l-A'_k),
$$
where $(A'_1,\dots,A'_{n-1})$ is what you get from $(A_1,\dots,A_n)$ by removing the entry corresponding to the deleted column.  By the proposition, this determinant is nonzero.  Thus the Jacobian matrix has full rank $n$ at every solution of Francesco's system.  Therefore every solution has multiplicity $1$.  
In summary, for every choice of $(m_1,\dots,m_n)$ in $P$, there are precisely $n!$ solutions of the original system.
Edit.  In fact, the argument above holds so long as $(m_1,\dots,m_n)$ is a sequence of complex numbers such that no "subset sum" equals $0$, i.e., $m_{i_1} + \dots + m_{i_q} \neq 0$ for every $1\leq i_1 < \dots < i_q \leq n$.  Since Francesco's homogeneous system defines a closed subscheme $Z$ of the product of projective spaces $\mathbb{P}^{n-1} \times \mathbb{P}^n$ with homogeneous coordinates $([m_1,\dots,m_n],[X_0,X_1,\dots,X_n])$, and since the projection $\pi_1:Z\to \mathbb{P}^{n-1}$ is finite, étale of degree $n!$ on the complement of the union of hyperplanes $Z(m_{i_1} + \dots + m_{i_q})$, then presumably the branch locus is precisely this union of hyperplanes.
