Summation of Powers over Product of Differences/Proof 1

Proof
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:

This is also seen to hold.

Basis for the Induction
$\map P 2$ is the case where $n = 2$:

When $0 \le r < n - 1$, it must be that $r = 0$:

When $r = n - 1 = 1$:

When $r = n = 2$:

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:

So:

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 $r = n$:

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:

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:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$