=== Definition 1 === :$V_n = \begin{bmatrix}   x_1       & \cdots & x_n \\   x_1^2     & \cdots & x_n^2 \\   \vdots    & \ddots & \vdots \\   x_1^{n}   & \cdots & x_n^{n} \\ \end{bmatrix} ,\quad V =  \begin{bmatrix}   1         & \cdots & 1 \\   x_1       & \cdots & x_n \\   \vdots    & \ddots & \vdots \\   x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{bmatrix} \quad $ Vandermonde matrices        :$D =  \begin{bmatrix}   x_1    & \cdots & 0 \\   \vdots & \ddots & \vdots \\   0      & \cdots & x_n \\ \end{bmatrix}, \quad P  =   \begin{bmatrix}   \map {p_1} {x_1} & \cdots & 0 \\   \vdots           & \ddots & \vdots \\   0                & \cdots & \map {p_n} {x_n} \\   \end{bmatrix} \quad $  Definition:Diagonal Matrix :$ E  =   \begin{bmatrix}   E_{11} & \cdots & E_{1n} \\   \vdots & \ddots & \vdots \\   E_{n1} & \cdots & E_{nn} \\   \end{bmatrix} \quad $ Matrix of symmetric functions :'''where for''' $\mathbf {1 \mathop \le i,j \mathop \le n}$: {{begin-eqn}} {{eqn | l = E_{ij}       | r =   \paren { -1 }^{n-j}   \,   e_{n-j}    \paren      {       \set {x_1,\ldots,x_n}        \setminus        \set {x_i}      }       | c = Definition:Elementary Symmetric Function $e_m \paren {U}$ }} {{eqn | l = \displaystyle  \map {p_j} x       | r = \prod_{ k \mathop = 1,\, k \mathop \neq j }^n  \paren { x - x_k } }} {{end-eqn}} === Lemma 1 === {{begin-eqn}} {{eqn | l = V_n       | r = V\, D       | c   Matrix Multiply }} {{eqn | l = V_n^{-1}       | r = D^{-1} V^{-1}       | c = provided the inverses exist: Inverse of Matrix Product }} {{end-eqn}} {{qed|lemma}} === Lemma 2 === :$ EV = P$ :$ V^{-1} = P^{-1} E\quad$ provided $\set {x_1,\ldots,x_n}$ is a set of distinct values. :$ V_n^{-1} = D^{-1} V^{-1}\quad$ provided $\set {x_1,\ldots,x_n}$ is a set of distinct values, all nonzero. '''Proof of Lemma 2''' Matrix multiply establishes $EV=P$, provided: :$(1)\quad \sum_{k = 1}^n E_{ik}\, x_j^{k-1} = \begin{cases} 0 & i \neq j \\ \map {p_i} {x_i} & i = j \end{cases} $   Polynomial $\map {p_i} {x}$ is zero for $x \in \set {x_1,\ldots,x_n} \setminus \set {x_i}$.                                  Then (1) is equivalent to :$(2)\quad \sum_{k = 1}^n  \paren { -1 }^{n-k}  e_{n-k}  \paren    {      \set {x_1,\ldots,x_n} \setminus \set {x_i}   }  \, x_j^{k-1}  =  \map {p_i} {x_j} $ Apply Viete's Formulas to degree $n-1$ monic polynomial $\map {p_i} {\mathbf {u} }$: :$(3)\quad  \sum_{k = 1}^{n} \paren {-1}^{n-k}  e_{n-k}  \paren    {      \set {x_1,\ldots,x_n} \setminus \set {x_i}   }  \, {\mathbf {u} }^{k-1} = \map {p_i} {\mathbf {u} }  $ Substitute ${\mathbf {u} } = x_j$ into (3), proving (2) holds. Then (2) and (1) hold, proving $EV=P$. Assume $\set {x_1,\ldots,x_n}$ is a set of distinct values. Then $\det \paren {P}$ is nonzero. By Matrix is Invertible iff Determinant has Multiplicative Inverse, $P$ has an inverse $P^{-1}$. Multiply $EV=P$ by $P^{-1}$, then:  :$V^{-1} = P^{-1} E\quad$ Left or Right Inverse of Matrix is Inverse Similarly, $D^{-1}$ exists provided $\set {x_1,\ldots,x_n}$ is a set of nonzero values. Then $V_n^{-1} = D^{-1} V^{-1}$ by Lemma 1.  {{qed|lemma}} '''Proof of the Theorem''' Assume $\set {x_1,\ldots,x_n}$ is a set of distinct values. Let $d_{ij}$ denote the entries in $V^{-1}$. Then: {{begin-eqn}} {{eqn | l = V^{-1}       | r = P^{-1} E       | c = Lemma 2 }} {{eqn | l = d_{ij}       | r = \dfrac{ E_{ij} }{\map {p_i} {x_i} }       | c = Matrix multiply }} {{eqn | r = \dfrac   { \displaystyle     \paren { -1 }^{n-j}     \,     e_{n-j}      \paren        {         \set {x_1,\ldots,x_n}          \setminus          \set {x_i}        }   }   { \displaystyle     \map {p_i} {x_i}    }       | c =Definition 1 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | n = 4       | l = d_{ij}       | r =   \paren { -1 }^{n-j}   \,   \dfrac   { \displaystyle     \sum_{            \substack              {                 1 \mathop \le m_1 \mathop \lt \cdots \mathop \lt m_{n-j} \mathop \le n \\                m_1,\ldots,m_{n-j} \mathop \neq i              }          }      x_{m_1} \cdots x_{m_{n-j} }   }   { \displaystyle     \prod_{              \substack                 {                  1 \mathop \le m \mathop \le n \\                  m \mathop \neq i                }           }      \paren { x_i - x_m }   }       | c = Definition:Elementary Symmetric Function $e_m \paren {U}$ Definition 1, equation for $\map {p_i} {x}$   }} {{end-eqn}}  Assume $\set {x_1,\ldots,x_n}$ is a set of distinct values, all nonzero. Let $b_{ij}$ denote the entries in $V_n^{-1}$. Then: {{begin-eqn}} {{eqn | l = V_n^{-1}       | r = D^{-1} V^{-1}       | c = Lemma 1 }} {{eqn | l = b_{ij}       | r = \dfrac{ 1 }{ x_i} \, d_{ij}       | c = Matrix multiply }} {{end-eqn}} Factor $\paren {-1}^{n-1}$ from the denominator of (4) to agree with Knuth (1997).  {{qed}}