=== Definition === {{begin-eqn}} {{eqn | l = \map p x        | r = \prod_{k \mathop = 1}^n \paren {x - x_k} }} {{eqn | l = \map {p_m} x       | r = \prod_{k \mathop = 1, \, k \mathop \ne m}^n \paren {x - x_k},\quad 1 \le m \le n }} {{eqn | l = A        | r = \begin{pmatrix}   1      & x_1    & \cdots & x_1^{n - 2} & x_1^{n - 1} \\   \vdots & \vdots & \cdots &   \vdots    & \vdots \\   1      & x_n    & \cdots & x_n^{n - 2} & x_n^{n - 1} \\   \end{pmatrix}       | c = Set $\set {x_1, \ldots, x_n}$ assumed to have distinct elements. }} {{eqn | l = A_{\vec Y}       | r = \begin{pmatrix}   1      & x_1    & \cdots & x_1^{n - 2} & y_1 \\   \vdots & \vdots & \cdots &   \vdots    & \vdots \\   1      & x_n    & \cdots & x_n^{n  -2} & y_n \\ \end{pmatrix}       | c = Vector $\vec Y$ has entries $y_1,\ldots,y_n$. }} {{end-eqn}} Symbol $\map {\mathbf {Cof } } {M, i, j}$ denotes cofactor $M_{i j}$ of matrix $M$ === Lemma 1 === {{begin-eqn}} {{eqn | l = \map \det A       | r = \map {p_m} {x_m} \map {\mathbf {Cof} } {A, m, n} }} {{end-eqn}} '''Proof of Lemma 1''' By Effect of Elementary Row Operations on Determinant, the determinant of $A$ is unchanged by adding a linear combination of the first $n - 1$ columns to the last column. Let $\map f x$ be the monic polynomial $\map {p_m} x = x^{n - 1} + $ lower power terms. Then: {{begin-eqn}} {{eqn | l = \map \det {\begin{pmatrix}   1      & x_1    & \cdots & x_1^{n - 2} & x_1^{n - 1} \\   \vdots & \vdots & \cdots &   \vdots    & \vdots \\   1      & x_n    & \cdots & x_n^{n - 2} & x_n^{n - 1} \\   \end{pmatrix} }       | r = \map \det {\begin{pmatrix}   1      & x_1    & \cdots & x_1^{n - 2} & \map f {x_1} \\   \vdots & \vdots & \cdots &   \vdots    & \vdots \\   1      & x_n    & \cdots & x_n^{n - 2} & \map f {x_n} \\  \end{pmatrix} } }} {{end-eqn}} The last column on the right has all components zero except $m$th entry $\map {p_m} {x_m}$. Apply cofactor expansion along column $n$. {{qed|lemma}} === Lemma 2 === {{begin-eqn}} {{eqn | n = 1       | l = \map \det A        | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i} }}    {{eqn | n = 2       | l = \map {\mathbf {Cof} } {A, m, n}       | r = \paren {-1}^{m + n} \prod_{1 \mathop \le i \mathop < j \mathop \le n, \, i \mathop \ne m, \, j \mathop \ne m} \paren {x_j - x_i}   }}    {{end-eqn}} '''Proof of Lemma 2''': Vandermonde Determinant establishes $(1)$. To prove (2), let: :$\set {y_1, \ldots, y_{n - 1} } = \set {x_1, \ldots, x_n} \setminus \set {x_m}$ then apply $(1)$ with $\set {x_1, \ldots, x_n}$ replaced by $\set {y_1, \ldots, y_{n - 1} }$. {{qed|lemma}} '''Theorem Details''': The Lemmas change the Theorem's equation to: {{begin-eqn}} {{eqn | n = 3       | l = \sum_{m \mathop = 1}^n {x_m^r} \, \dfrac {\map {\mathbf {Cof} } {A, m, n} } {\map \det A}       | r = \begin{cases} 0 & : 0 \mathop \le r \mathop < n - 1 \\ 1 & : r = n - 1 \\ \sum_{j \mathop = 1}^n x_j & : r = n \\     \end{cases} }} {{end-eqn}} Cofactor expansion changes identity $(3)$ to: {{begin-eqn}} {{eqn | n = 4       | l = \det \begin{pmatrix} 1      & x_1    & \cdots & x_1^{n - 2} & x_1^r  \\ \vdots & \vdots & \cdots &   \vdots    & \vdots \\ 1      & x_n    & \cdots & x_n^{n - 2} & x_n^r  \\     \end{pmatrix}       | r = \map \det A \begin{cases} 0 & : 0 \mathop \le r \mathop < n - 1 \\ 1 & : r = n - 1 \\ \sum_{j \mathop = 1}^n x_j & : r = n \\     \end{cases} }} {{end-eqn}} The proof is divided into three cases. '''Case 1''': $0 \mathop \le r \mathop \le n - 2$: The determinant on the left in $(4)$ is zero by Square Matrix with Duplicate Rows has Zero Determinant. '''Case 2''': $r = n - 1$: The determinant on the left in $(4)$ is $\map \det A$. '''Case 3''': $r = n$: Expand $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$ by Taylor's Theorem with remainder $R$ of degree $n - 2$: :$\map p x = x^n - \paren {x_1 + \cdots + x_n} x^{n - 1} + \map R x$ Let $x = x_k$ for $1 \mathop \le k \mathop \le n$. Then $x$ is a root of $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$: {{begin-eqn}} {{eqn | l = 0       | r = \map p {x_k} }} {{eqn | r = x_k^n - \paren {x_1 + \cdots + x_n} x_k^{n - 1} + \map R {x_k} }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = c       | r = x_1 + \cdots + x_n }} {{eqn | l = \vec u       | r = \begin {pmatrix} x_1^n \\ \vdots \\ x_n^n \end{pmatrix}       | c = Column $n$ of the left side of $(4)$ }} {{eqn | l = \vec z       | r = c \begin {pmatrix} x_1^{n - 1} \\ \vdots \\ x_n^{n - 1} \end{pmatrix}       | c = Constant $c$ times column $n$ of $A$ }} {{eqn | l = \vec w       | r = \begin{pmatrix} -\map R {x_1} \\ \vdots \\ -\map R {x_n} \end{pmatrix} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | n = 5       | l = \vec u       | r = \vec z + \vec w       | c = by equation $x_k^n = c x_k^{n - 1} - \map R {x_k}$ }} {{eqn | n = 6       | l = \map \det {A_{\vec z} }       | r = c \map \det A       | c = Effect of Elementary Row Operations on Determinant: factoring $c$ from last column of $A_{\vec z}$ }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 7       | l = \det \paren { A_{\vec w} }        | r = 0       | c = Effect of Elementary Row Operations on Determinant: $\vec w$ is a linear combination of columns $1$ to $n - 1$ }} {{end-eqn}} The left side of $(4)$: {{begin-eqn}} {{eqn | l = \map \det {A_{\vec u} }        | r = \map \det {A_{\vec z} } + \map \det {A_{\vec w} }       | c = by $(5)$ and Determinant as Sum of Determinants }} {{eqn | r = c \map \det A + 0       | c = by $(6)$ and $(7)$ }} {{eqn | r = \map \det A \displaystyle \sum_{i \mathop = 1}^n x_i       | c = definition of $c$ }} {{end-eqn}} {{qed}}