It's easy to see that for $Re(s)>0$,
$$\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{k=0}^{\infty}(-1)^k\frac{\ln^k(n)s^k}{k!}$$
from the series definition of the Riemann Zeta function and the Maclaurin series for $n^{-s}$ in terms of $s$. This gives the series representation of the coefficient for the $k$th term (i.e., the constant by which the term $s^k$ is scaled):
$$(-1)^k\frac{1}{k!}\displaystyle\sum_{n=1}^{\infty}\ln^k(n)$$
This begs the immediate question,
$$\displaystyle\sum_{n=1}^{\infty}\ln^k(n)=?$$
It should be noted that this series is divergent, but any way to analytically find an answer that would make sense would be great, like something along the lines of the method by which $\zeta(0)$ or the p-adic definition of $\sum_{n=0}^{\infty}2^n$ is found.
If you are not able to answer this question but are able to give me a representation of the Maclaurin series of the Riemann Zeta function without a double summation, that would be very nice too.