What is the Stirling transform of (k-1)?

AI Thread Summary
The discussion focuses on calculating the Stirling transform of (k-1)!, which involves using the explicit formula for the Stirling numbers of the second kind. The formula presented is complex, requiring a double summation that incorporates factorials and binomial coefficients. Mathematica simplifies this computation, yielding the result as (-1)^n Li_{1-n}(2). A link is provided to a proof that confirms this result, highlighting the mathematical intricacies involved in the Stirling transform. The conversation emphasizes the challenge of deriving the result manually and seeks further insights into the process.
Cristopher
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While reading about combinatorial mathematics, I came across the Stirling transform. https://en.wikipedia.org/wiki/Stirling_transform

So then, if I want to find the Stirling transform of, for instance, ##(k-1)!##, I have to compute this (using the explicit formula of the Stirling number of the second kind):

##\displaystyle\sum_{k=1}^{n}\left(\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n(k-1)!\right)##

This looks complicated and I don't know where to start. Mathematica gives the result ##(-1)^n\operatorname{Li}_{1-n}(2)##.

Any insights or hints of how to arrive at that result will be appreciated
 
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