Solving a multivariate polynomial system involving the power sums I would like to know if there is a way to solve or simplify the system of equations given by:
$$
x_1^1+x_2^1+\cdots x_n^1 = c_1\\
x_1^2+x_2^2+\cdots x_n^2 = c_2\\
\vdots\\
x_1^n+x_2^n+\cdots x_n^n = c_n
$$
 A: Use Newton's identities to recover the values of the symmetric functions 
$$ e_1(x_1,\ldots,x_n) = x_1+x_2+\ldots+x_n,\qquad e_2(x_1,\ldots,x_n) = \sum_{i<j}x_i x_j,\quad e_n(x_1,\ldots,x_n)=x_1\cdot x_2\cdot\ldots\cdot x_n$$
then $x_1,x_2,\ldots,x_n$ are the roots of the polynomial:
$$ p(x) = x^n - e_1 x^{n-1} + e_2 x^{n-2} -\ldots $$

We may also notice that
$$ r(x)= x^n p(1/x)=\prod_{i=1}^{n}(1- x_i x)=\sum_{k=0}^{n}x^{k}(-1)^k e_k $$
while:
$$ \frac{r'(x)}{r(x)}=\frac{d}{dx}\log r(x) = -\sum_{i=1}^{n}\frac{x_i}{1-x_i x} = -\sum_{i=1}^{n}\sum_{r\geq 0}x_i^{r+1}x^r=-\sum_{r\geq 0}c_{r+1}x^r $$
gives:
$$ r(x)=\exp\left(-\int\sum_{r\geq 0}c_{r+1}x^r\,dx\right) =\exp\left(-\sum_{r\geq 0}\frac{c_r}{r}\,x^r\right).$$
So we may compute the Taylor polynomial, up to the term $x^n$, of $\exp\left(-\sum_{r=1}^{n}\frac{c_r}{r}x^r\right)$, and by reversing its coefficients we have the monic polynomial $p(x)$ with roots $x_1,x_2,\ldots,x_n$. This approach is exactly equivalent to the previous one.
