Sum of Sequence of Binomial Coefficients by Sum of Powers of Integers

Theorem
Let $n, k \in \Z_{\ge 0}$ be positive integers.

Let $S_k = \ds \sum_{i \mathop = 1}^n i^k$.

Then:
 * $\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$

Proof
Let $N$ be a positive integer.

Then:

Summing from $N = 1$ to $N = n$, we have on the :

So, we have:

giving:


 * $\ds \sum_{i \mathop = 1}^k \binom {k + 1} i S_i = \paren {n + 1}^{k + 1} - \paren {n + 1}$