Binomial's Brother

Note that $\begin{aligned} F(k,\ell) & = \; \sum_{r=1}^{k+\ell}\frac{(-1)^{r-1}(k+r-1)!}{((r-1)!)^2(k-r+\ell)!} \\ & = \; \sum_{r=0}^{k+\ell-1} \frac{(-1)^r (k+r)!}{(r!)^2(k+\ell-1 - r)!} \; = \; \frac{1}{(k+\ell-1)!}\sum_{r=0}^{k+\ell-1}(-1)^r \binom{k+\ell-1}{r}\frac{(k+r)!}{r!} \\ & = \; \frac{1}{(k+\ell-1)!}\sum_{r=0}^{k+\ell-1} (-1)^r \binom{k+\ell-1}{r} \frac{d^k}{dx^k}x^{r+k} \Big|_{x=1} \; = \; \frac{1}{(k+\ell-1)!}\frac{d^k}{dx^k}\left(\sum_{r=0}^{k+\ell-1}(-1)^r\binom{k+\ell-1}{r}x^{r+k}\right)\Big|_{x=1} \\ & = \; \frac{1}{(k+\ell-1)!}\frac{d^k}{dx^k}\left(x^k(1-x)^{k+\ell-1}\right)\Big|_{x=1} \end{aligned}$ If $\ell \ge 2$ , then all terms in the $k$ th derivative of $x^k(1-x)^{k+\ell-1}$ will contain $1-x$ as a factor, and hence $F(k,\ell) \; = \; \sum_{r=1}^{k+\ell}\frac{(-1)^{r-1}(k+r-1)!}{((r-1)!)^2(k-r+\ell)!} \; =\; \boxed{0}$ which is the desired answer. However, there are some special cases which are worth noting. If $\ell=1$ then $F(k,1) \; = \; \frac{1}{k!}\frac{d^k}{dx^k}\left(x^k(1-x)^k\right)\Big|_{x=1} \; = \; \frac{1}{k!} \left(x^k \frac{d^k}{dx^k}(1-x)^k\right)\Big|_{x=1} \; = \; (-1)^k$ If $\ell=0$ then $F(k,0) \; = \; \frac{1}{(k-1)!}\frac{d^k}{dx^k}\left(x^k(1-x)^{k-1}\right)\Big|_{x=1} \; = \; \frac{1}{(k-1)!}\left(k \times \frac{d}{dx}x^k \times \frac{d^{k-1}}{dx^{k-1}}(1-x)^{k-1}\right)\Big|_{x=1} \; = \; (-1)^{k-1}k^2$