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

[Cristopher]

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

 

[Cristopher]

Just in case anyone is interested, today someone posted an excellent proof of why the Stirling transform of $(k-1)!$ is $(-1)^n \operatorname{Li}_{1-n}(2)$ here:

http://math.stackexchange.com/questions/1348593/stirling-transform-of-k-1