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

Proof
From the corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \left({-1}\right)^{r - k}$:


 * $\displaystyle \sum_k \binom r k \binom k n \left({-1}\right)^{r - k} = \delta_{n r}$

where $\delta_{n r}$ denotes the Kronecker delta.

Thus when $n \ne r$:
 * $\displaystyle \sum_k \binom r k \binom k n \left({-1}\right)^{r - k} = 0$

and so:
 * $\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} \left({c_0 \binom k 0 + c_1 \binom k 1 + \cdots + c_m \binom k m}\right) = c_r$

as the only term left standing is the $r$th one.

Choosing the coefficients $c_i$ as appropriate, a polynomial in $k$ can be expressed as a summation of binomial coefficients in the form:
 * $c_0 \dbinom k 0 + c_1 \dbinom k 1 + \cdots + c_m \dbinom k m$

Thus we can rewrite such a polynomial in $k$ as:
 * $b_0 + b_1 k + \cdots + b_r k^r$

{{explain|Why is the parameter of $b_r$ multiplied by $r!$?}

Hence the result.