Is it possible to simplify the following product of sums \begin{align} \prod_{k=1}^{l} \sum_{r=d}^m {{m}\choose{r}} \left(N-k \right)^{r} k^{m-r+1} \end{align} where $N>l$?
I was thinking that we can multiply and divide by $N^{m+1}$ to make it look more like a Binomial summation
\begin{align} \prod_{k=1}^{l} \sum_{r=d}^m \frac{N^{m+1}}{N^{m+1}} {{m}\choose{r}} \left(N-k \right)^{r} k^{m-r+1}= N^{r(m+1)}\prod_{k=1}^{l} \sum_{r=d}^m {{m}\choose{r}} \left(1-\frac{k}{N} \right)^{r} \left(\frac{k}{N} \right)^{m-r+1}. \end{align}