I would like to calculate $P(k) = \sum_{i=1}^{k} S_2(m,i) \frac{n!}{n^m (n-i)!}$ where $S_2(m,i)$ is a Stirling number of the second kind.
This problem is formulated from the question: uniformly draw $m$ samples with replacement from a set consists of $n$ unique elements, what is the probability that we get at most $k$ unique samples in the drawn $m$ samples?
The probelm is already answered here but I am interesting in if it could be simplified, or can we get a lower bound of the probability.
The probability you wrote is related to the development of powers into Falling Factorials $$ n^{\,m} = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,m} \right)} {S_2 (m,i){{n!} \over {\left( {n - i} \right)!}}} = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,m} \right)} {\left\{ \matrix{ m \cr i \cr} \right\}n^{\,\underline {\,i\,} } } $$ or $$ x^{\,m} = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,m} \right)} {\left\{ \matrix{ m \cr i \cr} \right\}x^{\,\underline {\,i\,} } } $$ where the curly brackets is another conventional way to write the Stirling N. of 2nd kind.
The Stirling N. (and falling factorials) satisfy many interesting relations, but as far as I am aware, none of them is useful to express the partial sum of the above expansion of $x^m$ into a form "closer" than the sum itself.
