Similar to what polynomials of $1/(1-q)$ does the function $\Gamma (n) (-\log (q))^{-n}$ behave near $q=1$? What polynomial of $\frac1{1-q}$ should we add to $\Gamma (n) (-\ln q)^{-n}$ so the result to be zero at $0$ and the pole to disappear?
 A: Below is an explicit formula that might be amenable to the OP's needs.  The 'polynomial family' can be given in terms of the Norlund polynomials, defined through the generating function 
$$ \Big( \frac{t}{e^t-1} \Big)^s e^{x\,t} = \sum_{k=0}^\infty B_k^{(s)}(x) \frac{t^k}{k!}  $$
In what follows, $B_k^{(s)}= B_k^{(s)}(0).$  Taking $s = -n$ and $t=\log(1-x)$ in the generating function means
$$ \Big( \frac{-x}{\log(1-x)} \Big)^n = \sum_{k=0}^\infty B_k^{(-n)} \frac{\log^k(1-x)}{k!}  $$
It is known, Digital Library Math. Func. 24.16.4
$$ \log^k(1-x)=k \,(-x)^k\sum_{m=0}^\infty \frac{B_m^{(m+k)}}{m+k} \frac{(-x)^m}{m!} $$
Insert this later expression into the penultimate and you have a double series.  Do a series rearrangement. (Hansen, A Table of Series of Products pg 14-15 has a slew of them.)  You'll get
$$ \Big( \frac{-x}{\log(1-x)} \Big)^n =1+\sum_{m=1}^\infty \frac{(-x)^m}{m} \sum_{k=1}^m 
\frac{B_k^{\,(-n)}}{(k-1)!} \frac{B_{m-k}^{\,(m)}}{(m-k)!}$$
These particular Norlund polys can be written, I believe, as Stirling numbers of the 1st and 2nd kind, but that obfuscates the derivation.
