Simplifying a sum of products related to Vandermonde determinant How to show this equality?
$$
1=(-1)^n\sum_{k=0}^n\frac{x_k^n}{\prod_{\substack{l=0 \\ l \neq k}}^n(x_l-x_k)}
$$
This is part of a proof to show the value of the determinant of the Vandermonde matrix equals some give product.
 A: Let 
\begin{align}
f(x)=\dfrac{(1-x_1x)(1-x_2x)\cdots(1-x_nx)}{(1-\dfrac{x_1}{x_0})(1-\dfrac{x_2}{x_0})\cdots(1-\dfrac{x_n}{x_0})}+\dfrac{(1-x_0x)(1-x_2x)\cdots(1-x_nx)}{(1-\dfrac{x_0}{x_1})(1-\dfrac{x_2}{x_1})\cdots(1-\dfrac{x_n}{x_1})}+\cdots+\dfrac{(1-x_0x)(1-x_1x)\cdots(1-x_{n-1}x)}{(1-\dfrac{x_0}{x_n})(1-\dfrac{x_1}{x_n})\cdots(1-\dfrac{x_{n-1}}{x_n})}
\\
\end{align}
$\hspace{17mm}=\sum \limits_{k=0}^{n}\prod \limits_{l=0,l\neq k}^{n}\dfrac{1-x_lx}{1-\dfrac{x_l}{x_k}}$
Now $\operatorname{deg}(f)=n$, and for each $0\leqslant k\leqslant n,\space f(\dfrac{1}{x_k})=1$. Since $f(x)-1$ is a polynomial of $n$, it only has $n$ roots. So $f(x)-1\equiv 0$ or $f(x)\equiv 1$. Take $x=0$, we have
$$f(0)=1=\sum \limits_{k=0}^{n}\prod \limits_{l=0,l\neq k}^{n}\dfrac{1}{1-\dfrac{x_l}{x_k}}=\sum \limits_{k=0}^{n}\prod \limits_{l=0,l\neq k}^{n}\dfrac{x_k^n}{x_k-x_l}=(-1)^n\sum \limits_{k=0}^{n}x_k^n\prod \limits_{l=0,l\neq k}^{n}\dfrac{1}{x_l-x_k}$$
A: Suppose we seek to verify that
$$1 = (-1)^n \sum_{k=0}^n 
\frac{x_k^n}{\prod_{l=0\atop l\ne k}^n (x_l - x_k)}$$
with the $x_l$ distinct.
This is the same as
$$1 = \sum_{k=0}^n 
\frac{x_k^n}{\prod_{l=0\atop l\ne k}^n (x_k - x_l)}.$$
Introduce $$f(z) = \frac{z^n}{\prod_{l=0}^n (z-x_l)}$$
which yields for the sum
$$\sum_{k=0}^n \mathrm{Res}_{z=x_k} f(z).$$
We  evaluate this using  the negative  of the  residue at  infinity as
these residues sum to zero. We obtain
$$\mathrm{Res}_{z=\infty} f(z)
= - \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1}{z^n}
\prod_{l=0}^n \frac{1}{1/z-x_l}
= - \mathrm{Res}_{z=0} \frac{1}{z^{n+2}}
\prod_{l=0}^n \frac{z}{1-x_l z}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z}
\prod_{l=0}^n \frac{1}{1-x_l z}.$$
This is the constant term i.e.
$$- [z^0] \prod_{l=0}^n \frac{1}{1-x_l z} = -1.$$
The negative of this is $$1$$
as claimed.
