$$S_{n}\equiv\sum_{j=1}^{N}x_{j}^{n}$$
If the values of $\{S_{1},S_{2},...,S_{N}\}$ are known, that places N constraints on the N unknowns $\{x_{1},x_{2},...,x_{N}\}$, so we could in theory calculate those values and use them to calculate all the higher $S_{n>N}$. However, that approach is impractical, and I don't need the $x_j$'s anyway.
Does there exist an expression for $S_{n}$ where $n>N$ in terms of $\{S_{1},S_{2},...,S_{N}\}$?
Examples:
If $N=1$, then $S_{n}=(S_{1})^{n}$ for all n.
If $N=2$ then $S_{3}=\frac{3}{2}S_{1}S_{2}-\frac{1}{2}S_{1}^{3}$ and $S_{4}=-\frac{1}{2}S_{1}^{4}+S_{2}S_{1}^{2}+\frac{1}{2}S_{2}^{2}$
Does a general expression in n and N exist, or do I need to write a program?
EDIT: This is related to Newton's Identities, but I did not manage to derive a general expression from them.