I am interested in a closed form of the following sum: $$ S_n := \sum _{i=1}^n\:\frac{\left(m-i\right)!}{\left(n-i\right)!}\;\; n,m \in \mathbb{N},\ n < m$$
Amongst other strategies I have also tried to compress the sum in order to gain some insight but failed in getting a useful result: $$ S_n = (m-n)\left(\frac{m-(n-1)}{1}\left(\frac{m-(n-2)}{2}\left(\cdots\left(\frac{m-1}{n-1} + 1\right)\cdots +1\right) + 1\right)+1\right)$$
I thought one may be able to express $S_n$ in terms of $S_{(n-1)}$ but, as far as I can see it, that makes things more complicated. This sum is used when analyzing the average runtime of a so called open addressing hash table.
We have $$\frac{(m-i)!}{(n-i)!}=\frac{(m-n)!(m-i)!}{(m-n)!(n-i)!}=(m-n)!\binom{m-i}{m-n}$$ so that $$\sum _{i=1}^n\:\frac{\left(m-i\right)!}{\left(n-i\right)!}=(m-n)!\sum_{k=m-n}^{m-1}\binom{k}{m-n}=(m-n)!\binom{m}{m-n+1}$$ by the hockey stick identity.