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For arbitrary integer $m, t, s<n$ prove the identity $$ \sum_{i=s+1}^{n}(i!)^2 \left\{ {n \atop i}\right\} \left\{ {m \atop i}\right\} \binom{s}{i}\binom{t}{i}=0, $$ here $\displaystyle \left\{ {n \atop i}\right\}$ is the Stirling numbers of the second kind.

I come across by accident to the identity as secondary result. I have checked it for many values, it is true. What about a proof in the general case?

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    $\begingroup$ Note that $\binom{s}{i}=0$ if $i>s$. $\endgroup$ – Markus Scheuer Jul 16 '18 at 16:45
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Since $s + 1 \leq i \leq n$, we see that $i>s \, \, \, \forall s < n$. Hence $\binom{s}{i}=0$. But each term of your sum is the product of $\binom{s}{i}$ and other values, which means each term will be zero. So the entire sum will be zero.

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