First consider the classical form of the Vandermonde matrix: :$W_n = \begin{bmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n - 1} \\   1 & x_2 & x_2^2 & \cdots & x_2^{n - 1} \\ \vdots & \vdots & \ddots & \vdots \\   1 & x_n & x_n^2 & \cdots & x_n^{n - 1} \\ \end{bmatrix}$ By Vandermonde Determinant, the determinant of $W_n$ is: :$\displaystyle \map \det {W_n} = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j} \ne 0$ Since this is non-zero, by Matrix is Invertible iff Determinant has Multiplicative Inverse, the inverse matrix, denoted $B = \sqbrk {b_{i j} }$, is guaranteed to exist. Using the definition of the matrix product and the inverse: :$\displaystyle \sum_{k \mathop = 1}^n b_{k j} x_i^{k - 1} = \delta_{i j}$ That is, if $\map {P_j} x$ is the polynomial: :$\displaystyle \map {P_j} x := \sum_{k \mathop = 1}^n b_{k j}x^{k - 1}$ then: :$\map {P_j} {x_1} = 0, \ldots, \map {P_j} {x_{j - 1} } = 0, \map {P_j} {x_j} = 1, \map {P_j} {x_{j + 1} } = 0, \ldots, \map {P_j} {x_n} = 0$ By the Lagrange Interpolation Formula, the $j$th row of $B$ is composed of the coefficients of the $j$th Lagrange basis polynomial: :$\displaystyle \map {P_j} x = \sum_{k \mathop = 1}^n b_{k j} x^{k - 1} = \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \frac {x - x_m} {x_j - x_m}$ Identifying the $k$th order coefficient in these two polynomials yields: :$b_{k j} = \begin{cases} \paren {-1}^{n - k} \paren {\dfrac {\displaystyle \sum_{\substack {1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n - k} \mathop \le n \\ m_1, \ldots, m_{n - k} \mathop \ne j} } x_{m_1} \cdots x_{m_{n - k} } } {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \paren {x_j - x_m} } } & : 1 \le k < n \\ \qquad \qquad \qquad \dfrac 1 {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \paren {x_j - x_m}} & : k = n \end{cases}$ which gives: :$b_{k j} = \begin{cases} \paren {-1}^{k - 1} \paren {\dfrac {\displaystyle \sum_{\substack {1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n - k} \mathop \le n \\ m_1, \ldots, m_{n - k} \mathop \ne j} } x_{m_1} \cdots x_{m_{n - k} } } {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \paren {x_m - x_j} } } & : 1 \le k < n \\ \qquad \qquad \qquad \dfrac 1 {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \paren {x_j - x_m} } & : k = n \end{cases}$ For the general case, we observe that by simple multiplication: :$\displaystyle V_n =  \begin {pmatrix} \begin {bmatrix}   x_1 & 0 & \cdots & 0 \\   0 & x_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\   0 & 0 & \cdots & x_n \end {bmatrix} \cdot W_n \end {pmatrix}^\intercal$ So by Inverse of Matrix Product and Inverse of Diagonal Matrix: :$\displaystyle V_n^{-1} =  \begin {bmatrix}   x_1^{-1} & 0 & \cdots & 0 \\   0 & x_2^{-1} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\   0 & 0 & \cdots & x_n^{-1} \end {bmatrix} \cdot \paren {W_n^{-1} }^\intercal$ Let $c_{k j}$ denote the $\tuple {k, j}$th coefficient of $V_n^{-1}$. Since the first matrix in the product expression for $V_n^{-1}$ above is diagonal: :$c_{kj} = \dfrac 1 {x_k} b_{j k}$ which establishes the result. {{Qed}}