How to find a polynomial expression for $\ln(x)^{i}$, where $x>0$, and $i \in \mathbb{N}^{*} \setminus \{1\}$?

Question

Seems a bit difficult since, even if it possible to say:

$\displaystyle \forall \,\, 0 < a < 1,\,\, \ln(a) \,= \sum \limits_{n=1}^{+\infty} (-1)^{n+1} \frac{(a-1)^n}{n}$

$\displaystyle \forall \,\, a > 1,\,\, \ln(a) \,= \sum \limits_{n=1}^{+\infty} \frac{1}{n} \left(\frac{a-1}{a}\right)^n$

If you power up the logarithm, it may not end in an infinite Newton power sum expression development, since the power series expressing the logarithm is an infinite sum...

There must be other ways.

Answer 1

It looks to me like $$\ln(x)^i = i! \sum_{n=i}^\infty s(n,i) \frac{(x-1)^n}{n!} $$ for positive integers $i$ and $|x-1|<1$, where $s(n,i)$ are (signed) Stirling numbers of the first kind.

EDIT: Then also

$$ \ln(x)^i = (-1)^{n+i} \ln(1/x)^i = (-1)^i i! \sum_{n=i}^\infty \frac{s(n,i)}{n!} \left( \frac{x-1}{x}\right)^n $$ for $|(x-1)/x| < 1$, which covers $x > 1$.