Sum of Elements in Inverse of Vandermonde Matrix

Theorem
Let $V_n$ be the Vandermonde matrix of order $n$ given by:


 * $V_n = \begin{bmatrix}

x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^n & x_2^n & \cdots & x_n^n \end{bmatrix}$

Let $V_n^{-1}$ be its inverse, from Inverse of Vandermonde Matrix:
 * $b_{i j} = \begin{cases}

\left({-1}\right)^{j-1} \left({\dfrac{\displaystyle \sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-j} \mathop \le n \\ m_1, \ldots, m_{n-j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n-j}} } {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_m - x_i}\right)}}\right) & : 1 \le j < n \\ \qquad \qquad \qquad \dfrac 1 {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \left({x_i - x_m}\right)} & : j = n \end{cases}$

The sum of all the elements of $V_n^{-1}$ is:
 * $\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = 1 - \prod_{k \mathop = 1}^n \left({1 - \dfrac 1 {x_k} }\right)$

Proof
< GB Gustafson 2019 -> Apply Sum of Elements of Invertible Matrix to $B=V_n^{-1}$.