Girard–Newton formulae，亦称牛顿恒等式
不失一般性，设$n$次多项式：
$$
\begin{align}
P(x) &= x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_0\\
&= (x-x_1)(x-x_2)...(x-x_{n-1})(x-x_{n})
\end{align}
$$
定义：
$$
\begin{align}
p_i &= \sum_{j=1}^{n}x_j^i \\
&= x_1^i+x_2^i+...+x_{n-1}^i+x_{n}^i
\end{align}
$$
有韦达定理：
$$
\begin{cases}
e_0 = 1\\
e_1 = \displaystyle \sum_i x_i = -a_{n-1}\\
e_2 = \displaystyle \sum_{i < j} x_ix_j = a_{n-2}\\
e_3 = \displaystyle \sum_{i < j < k} x_ix_jx_k = -a_{n-2}\\ ...\\ e_n = x_ix_j...x_n = (-1)^n{a_{0}}\\ \text{let }e_{n+1} = e_{n+2} = ... =0 \end{cases} $$ 则有以下恒等式： $$ \begin{cases} p_0 &= \displaystyle \sum_{j=1}^{n}x_j^0 = n\\ p_1 &= \displaystyle \sum_{j=1}^{n}x_j^1 = e_1\\ p_2 &= \displaystyle \sum_{j=1}^{n}x_j^2 = e_1p_1 - e_2 \times 2\\ p_3 &= \displaystyle \sum_{j=1}^{n}x_j^3 = e_1p_2 - e_2p_1 + e_3 \times 3\\ ...\\ p_n &= \displaystyle \sum_{j=1}^{n}x_j^n = e_1p_{n-1} - e_2p_{n-2} + ... + (-1)^{n-2}e_{n-1}p_1 + (-1)^{n-1}e_{n} \times n \end{cases} $$ 即对$m \le n$时，有： $$ p_m = \sum_{j=1}^{n}x_j^m = e_1p_{m-1} - e_2p_{m-2} + ... + (-1)^{m-2}e_{m-1}p_1 + (-1)^{m-1}e_{m} \times m $$ 亦或者： $$ me_m = e_{m-1}p_1 - e_{m-2}p_2 + ... + (-1)^{m-1}e_0p_m $$ 当$m > n$时，则有：
$$ p_m = \sum_{j=1}^{n}x_j^m = e_1p_{m-1} - e_2p_{m-2} + ... + (-1)^{n-1}e_{n}p_{m-n} $$
证明：
(Ⅰ)当$m=n$时，设$x_1x_2...x_n$为$P(x)=0$的$n$个根，而
$$
\begin{align}
P(x) &= x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_0\\
&= x^n-e_1x^{n-1}+...+(-1)^n{e_n}
\end{align}
$$
将$x_1x_2...x_n$分别带入$P(x)=0$可得方程组：
$$
\begin{cases}
P(x_1) &= x_1^n-e_1x_1^{n-1}+...+(-1)^n{e_n} &= 0\\
P(x_2) &= x_2^n-e_1x_2^{n-1}+...+(-1)^n{e_n} &= 0\\
...\\
P(x_n) &= x_n^n-e_1x_n^{n-1}+...+(-1)^n{e_n} &= 0\\
\end{cases}
$$
将方程组全加起来，可得：
$$
(x_1^n+x_2^n+...+x_n^n)-e_1(x_1^{n-1}+x_2^{n-1}+...+x_n^{n-1})+...+n(-1)^ne_n=0
$$
即：
$$
p_n-e_1p_{n-1}+...+(-1)^ne_nn=0
$$
或者
$$
p_n-e_1p_{n-1}+...+(-1)^ne_np_0=0
$$
(Ⅱ)当$m>n$时,令
$$
\begin{align}
Q(x) &= P(x)x^{m-n} \\
&= x^{m}+a_{n-1}x^{m-1}+...+a_{1}x^{m-n+1}+a_0x^{m-n} \\
&= (x-x_1)(x-x_2)...(x-x_{n-1})(x-x_{n})x^{m-n}
\end{align}
$$
则方程$Q(x)=0$相比$P(x)=0$多$m-n$个值为$0$的根
设$Q(x)=0$的根的$i$次方之和为$p'_i$：
$$
\begin{align}
p'_i &= \sum_{j=1}^{m}{x'}_j^i \\
&= \sum_{j=1}^{n}x_j^i \\
&= p_i
\end{align}
$$
$$
\begin{align}
e'_k &= \displaystyle \sum_{i_1 < i_2 < ... < i_k} x_{i_1}x_{i_2}...x_{i_k} \\
&=
\begin{cases}
0, & (n < k \le m)\\ e_k, &(k \le n) \end{cases}\\ &= e_k \end{align} $$ 同(Ⅰ)可以得到： $$ p'_m-e'_1p'_{m-1}+...+(-1)^me'_mp'_0=0 $$ 亦即： $$ p_m-e_1p_{m-1}+...+(-1)^me_mp_0=0 $$ $\because e_k=0 (k>n)$，可以删除后$m-n$项：
$$
p_m-e_1p_{m-1}+...+(-1)^ne_np_{m-n}=0
$$
(Ⅲ)当$m < n$时，方程
$$
\begin{align}
P(x) &= x^n-e_1x^{n-1}+...+(-1)^n{e_n}\\
&= (x-x_1)(x-x_2)...(x-x_{n-1})(x-x_{n})
\end{align}
$$
两边同时对$x$求导：
$$
\begin{align}
&nx^{n-1}-(n-1)e_1x^{n-2}+...+(-1)^{n-1}e_{n-1} \\
&= \frac{P(x)}{x-x_1}+\frac{P(x)}{x-x_2}+...+\frac{P(x)}{x-x_n} = \sum_k \frac{P(x)}{x-x_k} \tag{1}
\end{align}
$$
注意到：
$$
\begin{align}
\frac{x^{m+1}P(x)}{x-x_k} &= \frac{(x^{m+1}-x_k^{m+1})P(x)}{x-x_k}+\frac{x_k^{m+1}P(x)}{x-x_k} \\
&= \frac{(x-x_k)(x^m+x^{m-1}x_k^1+...+x_k^m)P(x)}{x-x_k}+\frac{x_k^{m+1}P(x)}{x-x_k} \\
&= (x^m+x^{m-1}x_k^1+...+x_k^m)P(x)+\frac{x_k^{m+1}P(x)}{x-x_k}
\end{align} \tag{2}
$$
将$(1)$式两边同时乘以$x^{m+1}$，并将$(2)$式代入：
$$
\begin{align}
& x^{m+1}(nx^{n-1}-(n-1)e_1x^{n-2}+...+(-1)^{n-1}e_{n-1}) \\
&= x^{m+1}\sum_k \frac{P(x)}{x-x_k} \\
&= \sum_k \frac{x^{m+1}P(x)}{x-x_k} \\
&= \sum_k ((x^m+x^{m-1}x_k^1+...+x_k^m)P(x)+\frac{x_k^{m+1}P(x)}{x-x_k}) \\
\end{align}
$$
即：
$$
\begin{align}
&P(x)\sum_k (x^m+x^{m-1}x_k^1+...+x_k^m)+\sum_k \frac{x_k^{m+1}P(x)}{x-x_k} \\
&= x^{m+1}(nx^{n-1}-(n-1)e_1x^{n-2}+...+(-1)^{n-1}e_{n-1}) \tag{3}
\end{align}
$$
其中：
$$
\begin{align}
&\sum_k (x^m+x^{m-1}x_k^1+...+x_k^m) \\
&= \sum_k x^m+\sum_k x^{m-1}x_k^1+...+\sum_k x_k^m\\
&= nx^m+p_1x^{m-1}+...+p_m \tag{4}
\end{align}
$$
将$P(x)$表达式及$(4)$式代入$(3)$式：
$$
\begin{align}
&(x^n-e_1x^{n-1}+...+(-1)^n{e_n})(nx^m+p_1x^{m-1}+...+p_m)+\sum_k \frac{x_k^{m+1}P(x)}{x-x_k} \\
&= x^{m+1}(nx^{n-1}-(n-1)e_1x^{n-2}+...+(-1)^{n-1}e_{n-1}) \tag{5}
\end{align}
$$
利用$(5)$式左右两边关于$x^n$的系数相等：
$$
p_m-e_1p_{m-1}+...+(-1)^{m-1}e_{m-1}p_1+(-1)^me_mn = (-1)^m(n-m)e_m
$$
化简得：
$$
\begin{align}
p_m-e_1p_{m-1}+...+(-1)^{m-1}e_{m-1}p_1+(-1)^mme_m = 0 & \\
\blacksquare &
\end{align}
$$
参考文档
https://zhuanlan.zhihu.com/p/146243974
https://www.hanspub.org/journal/PaperInformation?paperID=49744