# Proving an identiy on elementary symmetric polynomials

I happened to meet the following exercise on elementary symmetric polynomials. Here is the wiki page for the definition of elementary symmetric polynomial.

Let $$n\ge 3$$ and $$e_k:=e_k(x_1, \cdots, x_n)$$ be the $$k$$-th elementary symmetric polynomial with respect to variables $$x_1, \cdots, x_n$$. Prove the following identity: $$\sum_{i=2}^n\frac{e_n(x_1-x_i)}{x_i^2 x_1}+2e_{n-2}=x_1\frac{e_{n-1}}{e_n}e_{n-2}(x|x_1)$$ where $$e_{n-2}(x|x_1)=e_{n-2}-x_1 e_{n-3}(x_2,\cdots, x_n)$$

I have computed precisely for the case $$n=3$$ and $$n=4$$, and have verified the identity. However, I have trouble in proving the identity for general $$n$$. Can anyone present an elegant proof? Thanks very much in advance.

Let $$e_k$$ be the elementary symmetric polynomials in the $$n$$ variables $$(x_1,x_2,\dots,x_n)$$, and let $$d_k$$ be the elementary symmetric polynomials in the $$n-1$$ variables $$(x_2,x_3,\dots,x_n)\,$$. It follows from the definitions that $$\,e_n=x_1d_{n-1}\,$$ and $$\,e_k=d_k+x_1d_{k-1}\,$$ for $$\,1 \le k \le n-1\,$$.
In particular $$\,e_{n-2}(x|x_1)=e_{n-2}-x_1 e_{n-3}(x_2,\cdots, x_n)=e_{n-2}-x_1d_{n-3}=d_{n-2}\,$$, so the identity to prove can be rewritten as:
\require{cancel} \begin{align} \sum_{i=2}^n\frac{e_n(x_1-x_i)}{x_i^2 x_1}+2e_{n-2}&=x_1\frac{e_{n-1}}{e_n}e_{n-2}(x|x_1) \\ \iff\;\;x_1d_{n-1} \left(\sum_{i=2}^n\frac{1}{x_i^2}-\frac{1}{x_1}\sum_{i=2}^n\frac{1}{x_i}\right)+2\left(d_{n-2}+x_1d_{n-3}\right)&=\bcancel{x_1}\frac{d_{n-1}+x_1d_{n-2}}{\bcancel{x_1}d_{n-1}}d_{n-2} \end{align}
• $$\displaystyle \sum_{i=2}^n \frac{1}{x_i} = \frac{d_{n-2}}{d_{n-1}}$$
• $$\displaystyle \sum_{i=2}^n \frac{1}{x_i^2} = \left(\frac{d_{n-2}}{d_{n-1}}\right)^2-2 \frac{d_{n-3}}{d_{n-1}}$$