proof of $\sum\nolimits_{i = 1}^{n } {\prod\nolimits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i – x_j }}} } = 1$

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  • Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$

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Consider the Lagrange interpolation of the polynomial $f(x)=x^{n-1}$ with interpolation points $x_1,\ldots,x_n$. We have:
$$x^{n-1}=\sum_{i=1}^{n}x_i^{n-1}\prod_{j\neq i}\frac{x-x_j}{x_i-x_j},$$
so, by comparing the leading coefficients of RHS and LHS, the result follows.

For $n>1$, the partial fractional decomposition of
$$
\frac1z\prod_{j=1}^{n-1}\frac{z}{z-x_j}
=\frac{z^{n-2}}{\prod\limits_{j=1}^{n-1}(z-x_j)}
=\sum_{i=1}^{n-1}\frac{A_i}{z-x_i}\tag{1}
$$
using the Heaviside method yields
$$
A_i=\frac{x_i^{n-2}}{\prod\limits_{\substack{j=1\\j\ne i}}^{n-1}(x_i-x_j)}
=\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{2}
$$
Combine $(1)$ and $(2)$:
$$
\begin{align}
\prod_{j=1}^{n-1}\frac{z}{z-x_j}
&=\sum_{i=1}^{n-1}A_i\frac{z}{z-x_i}\\
&=\sum_{i=1}^{n-1}\frac{z}{z-x_i}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\\
&=\sum_{i=1}^{n-1}\left(1-\frac{x_i}{x_i-z}\right)\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{3}
\end{align}
$$
Set $z=x_n$ in $(3)$:
$$
\prod_{j=1}^{n-1}\frac{x_n}{x_n-x_j}
=\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}
-\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}\tag{4}
$$
Add the second term of the right side of $(4)$ to both sides:
$$
\sum_{i=1}^n\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}
=\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{5}
$$
Noting that the case $n=1$ follows vacuously, using $(5)$ and induction proves that
$$
\sum_{i=1}^n\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}=1\tag{6}
$$