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Assume $F$ is a field of zero characteristic. Denote the elementary symmetric polynomials of $n$ variables by $e_k$, $\quad k=\overline{1,n}$. Let the symbol $\sum ax_1^{i_1}\dots x_n^{i_n}$ denote the sum of all different monomials which can be obtained from $ax_1^{i_1}\dots x_n^{i_n}$ by permuting its variables. For instance, if $n=3$,

$$\sum x_1^2x_2=x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2$$

The polynomials $S_k=\sum_{i=1}^n x_i^k$ are called power sum symmetric polynomials. Obviously $S_1=e_1$ and $S_2=e_1^2-2e_2$. In order to prove Newton's identities one can see that for $2<k<n+1$

$$e_iS_{k-i}=\sum x_1^{k-i+1}x_2\dots x_i+\sum x_1^{k-i}x_2\dots x_ix_{i+1}$$

How can we see the above identity?

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  • $\begingroup$ Does it help to consider the relation $1 + e_1 + \dots + e_n = (1+x_1)(1+x_2)\dots(1+x_n)$? $\endgroup$ – bryanj Jun 23 '14 at 12:30
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EDIT: I don't think I answered the precise question that you asked, but instead sketched another proof of Newton's relations.

If you're talking about this,
then consider the relation $1 + e_1 + \dots + e_n = (1+x_1)(1+x_2)\dots(1+x_n)$.

For an indeterminate $t$, you get:
$t^n + t^{n-1} e_1 + \dots + t^0 e_n = (t+x_1)(t+x_2)\dots(t+x_n)$

Substitute $t = -x_k$:
$(-1)^n x_k^n + (-1)^{n-1} x_k^{n-1} e_1 + \dots + e_n = 0$

Do this for all $x_k$, $k = 1, \dots, n$ and sum:

$(-1)^n (x_1^n + \dots + x_n^n) + (-1)^{n-1} (x_1^{n-1} + \dots + x_n^{n-1})e_1 + \dots + n e_n = 0$

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