The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle S_n := \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ === Edge Cases === $\map P 0$ is a vacuous summation: :$\displaystyle S_0 := \sum_{j \mathop = 1}^0 \paren {\dfrac { {x_j}^0} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 0 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = 0 = \sum_{j \mathop = 1}^0 x_0$ which is seen to hold. $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^1 \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | r = \dfrac { {x_1}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 1 \\ k \mathop \ne 1} } \paren {x_1 - x_k} }       | c =  }} {{eqn | r = \dfrac { {x_1}^r} 1       | c = as the product is vacuous }} {{eqn | r = {x_1}^r       | c = simplification }} {{eqn | r = \begin{cases} 1 & : r = 0 \\ x_1 & : r = 1 \end{cases}       | c =  }} {{end-eqn}} This is also seen to hold. === Basis for the Induction === $\map P 2$ is the case where $n = 2$: {{begin-eqn}} {{eqn | l = S_2       | r = \sum_{j \mathop = 1}^2 \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 2 \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | c =  }} {{eqn | r = \frac { {x_1}^r} {x_1 - x_2} + \frac { {x_2}^r} {x_2 - x_1}       | c =  }} {{eqn | r = \frac { {x_1}^r - {x_2}^r} {x_1 - x_2}       | c =  }} {{end-eqn}} When $0 \le r < n - 1$, it must be that $r = 0$: {{begin-eqn}} {{eqn | l = S_2       | r = \frac {1 - 1} {x_1 - x_2}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} When $r = n - 1 = 1$: {{begin-eqn}} {{eqn | l = S_2       | r = \frac {x_1 - x_2} {x_1 - x_2}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} When $r = n = 2$: {{begin-eqn}} {{eqn | l = S_2       | r = \frac { {x_1}^2 - {x_2}^2} {x_1 - x_2}       | c =  }} {{eqn | r = \frac {\paren {x_1 + x_2} \paren {x_1 - x_2} } {x_1 - x_2}       | c =  }} {{eqn | r = x_1 + x_2       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^2 x_j       | c =  }} {{end-eqn}} Thus in all cases, $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^m \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le m \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < m - 1 \\ 1 & : r = m - 1 \\ \displaystyle \sum_{j \mathop = 1}^m x_j & : r = m \end{cases}$ from which it is to be shown that: :$\displaystyle \sum_{j \mathop = 1}^{m + 1} \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le m + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < m \\ 1 & : r = m \\ \displaystyle \sum_{j \mathop = 1}^{m + 1} x_j & : r = {m + 1} \end{cases}$ === Induction Step === This is the induction step: For $n > 2$, let the formula be rewritten: {{begin-eqn}} {{eqn | l = S_n       | r = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | c =  }} {{eqn | r = \frac 1 {x_n - x_{n - 1} } \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_n - x_{n - 1} } } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | c =  }} {{eqn | r = \frac 1 {x_n - x_{n - 1} } \paren {\sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_j - x_n} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_j - x_{n - 1} } } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = S_{n + 1}       | r = \frac 1 {x_{n + 1} - x_n} \paren {\sum_{j \mathop = 1}^{n + 1} \paren {\dfrac { {x_j}^r \paren {x_j - x_{n + 1} } } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^{n + 1} \paren {\dfrac { {x_j}^r \paren {x_j - x_n} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } }       | c =  }} {{eqn | r = \frac 1 {x_{n + 1} - x_n} \paren {\sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } \paren {\dfrac { {x_j}^r } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j \\ k \mathop \ne n} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }       | c = as both parts are the original sum }} {{end-eqn}} When $r < n$, both parts are equal to $0$ or $1$ by the induction hypothesis. Thus either: :$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {1 - 1} = 0$ or: :$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {0 - 0} = 0$ and so $\map P {m + 1}$ holds for $r < n$. When $r = n$: {{begin-eqn}} {{eqn | l = \sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } \paren {\dfrac { {x_j}^r } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j \\ k \mathop \ne n} } \paren {x_j - x_k} } }       | r = \sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } x_j       | c =  }} {{eqn | l = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | r = \sum_{j \mathop = 1}^n x_j       | c =  }} {{end-eqn}} Thus: :$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {x_{n + 1} - x_n} = 1$ and so $\map P {m + 1}$ holds for $r = n$. Now: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^n \paren {\dfrac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_j - x_k} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | r = 0       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^n - \paren {x_1 + \cdots + x_n} {x_j}^{n + 1} + \map P {x_j} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | c = where $\map P {x_j}$ is a polynomial of degree $n - 2$ }} {{eqn | ll= \leadsto       | l = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^n} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }       | r = \sum_{l \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n x_l \paren {\dfrac { {x_j}^{n - 1} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }       | c =  }} {{eqn | r = \sum_{l \mathop = 1}^n x_l       | c =  }} {{end-eqn}} Thus $\map P {m + 1}$ holds for $r = n + 1$. So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ {{qed}}