I entered this series into Mathematica to see if it could be simplified and it managed to give a form in terms of the partial gamma function. However I do not know how it derives this formula and I am interested in knowing how something like that can be done.
$$\sum _{k=0}^n \frac{\log ^{n-k}(x)}{(n-1)^{k+1} x^{n-1} (n-k)!}=\frac{\;\Gamma (n+1,\;(n-1)\log x) }{(n-1)^{n+1}\;n!}$$
I presume that the following series is used $$\Gamma(s,x) = (s-1)!\, e^{-x} \sum_{k=0}^{s-1} \frac{x^k}{k!}$$ Where $s$ is an integer.
However after some manipulation I could not even get the orignal sum from the answer.