Find series expansion of $\log^{\alpha}(1-x)$

Question

We know that $$\log(1-x) = - \sum^{\infty}_{n=1} \frac{x^n}n; \quad 0<x<1.$$ I would like to know what the series expansion of $$\log^{\alpha}(1-x)=?\, \mbox{or}\, \log^{\alpha}(\frac{1}{1-x})=?\,; \, \mbox{for}\quad 0<x<1 \, \mbox{and}\, -1<\alpha<0.$$

Thank you in advance

Answer 1

https://de.wikipedia.org/wiki/Stirling-Zahl look for "Stirling-Polynome" (Stirling polynomials)
you have to scroll down a bit, it's $\displaystyle(\frac{-\ln(1-t)}{t})^x$