Sum over k of r Choose k by -1^r-k by Polynomial/Proof 2

Proof
From Summation of Powers over Product of Differences:
 * $\displaystyle \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$

Now we have: