Summation of Powers over Product of Differences/Proof 1

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle S_n := \sum_{j \mathop = 1}^n \left({\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} }\right) = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$

Edge Cases
$P \left({0}\right)$ is a vacuous summation:


 * $\displaystyle S_0 := \sum_{j \mathop = 1}^0 \left({\dfrac { {x_j}^0} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 0 \\ k \mathop \ne j} } \left({x_j - x_k}\right)} }\right) = 0 = \sum_{j \mathop = 1}^0 x_0$

which is seen to hold.

$P \left({1}\right)$ is the case:

This is also seen to hold.

Basis for the Induction
$P \left({2}\right)$ 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, $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 1}^m \left({\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le m \\ k \mathop \ne j} } \left({x_j - x_k}\right)} }\right) = \begin{cases} 0 & : 0 \le r < m - 1 \\ 1 & : r = m - 1 \\ \displaystyle \sum_{j \mathop = 1}^m x_j & : r = m \end{cases}$

from which it is to be shown that:
 * $\displaystyle \sum_{j \mathop = 1}^{m + 1} \left({\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le m + 1 \\ k \mathop \ne j} } \left({x_j - x_k}\right)} }\right) = \begin{cases} 0 & : 0 \le r < m \\ 1 & : r = m \\ \displaystyle \sum_{j \mathop = 1}^{m + 1} x_j & : r = {m + 1} \end{cases}$

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:
 * $S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \left({1 - 1}\right) = 0$

or:
 * $S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \left({0 - 0}\right) = 0$

and so $P \left({m + 1}\right)$ holds for $r < n$.

When $r = n$:

Thus:
 * $S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \left({x_{n + 1} - x_n}\right) = 1$

and so $P \left({m + 1}\right)$ holds for $r = n$.

Now:

Thus $P \left({m + 1}\right)$ holds for $r = n + 1$.

So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

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