I evaluated the integral you mentioned above and arrived at $\int_{0}^{1}t^{m}ln^{n}(t)dt=(-1)^{n}\frac{n!}{(m+1)^{n+1}}$.
May I ask how this is used to evaluate the problem at hand?. I tried using $(1-t)^{m}$ instead of $t^{n}$ letting $m=-n$. But it turned nasty when I used parts as I done with the first part. I am certain there is something I am not seeing. I did manage to get the $(-1)^{n}\cdot n$ portion as in RobJohn's solution, but not the Stirling/zeta portion
Here's how I managed the integral you mentioned. I just do not know how to relate it to the one I posted.
$\int_{0}^{1}t^{m}ln^{n}(t)dt$
Use parts and get:
$\frac{t^{m+1}}{m+1}ln^{n}(t)-\frac{n}{m+1}\int_{0}^{1}t^{m}ln^{n}(t)dt$
Using the limits gives:
$\frac{-n}{m+1}\int_{0}^{1}t^{m}ln^{n-1}(t)dt$........[1]
Change n to n-1:
$\int_{0}^{1}t^{m}ln^{n-1}(t)dt=-\frac{n-1}{m+1}\int_{0}^{1}t^{m}ln^{n-2}(t)dt$
Sub this into [1]:
$(-1)^{2}\frac{n(n-1)}{(m+1)^{2}}\int_{0}^{1}t^{m}ln^{n-2}(t)dt$
Now, continue repeating and generally we have:
$\int_{0}^{1}t^{m}ln^{n}(t)dt=(-1)^{n}\frac{n!}{(m+1)^{n}}\int_{0}^{1}x^{m}dx$
$=(-1)^{n}\frac{n!}{(m+1)^{n+1}}$
Now, how can I relate to the Stirling and zeta to arrive at the closed form RobJohn showed?.
I tried the same sort of method and let $m=-n$. This kind of threw a wrench in the whole mess. I managed to see the $(-1)^{n}\cdot n$, but not the stirling/zeta portion.
Using parts: $dv=(1-t)^{-n}dt, \;\ u=ln^{n}(t)dt, \;\ du=\frac{nln^{n-1}(t)}{t}dt$
$v=\frac{1}{(1-t)^{n-1}(n-1)}$
This leads to $-\frac{n}{n-1}\int_{0}^{1}\frac{ln^{n-1}(t)}{t(1-t)^{n-1}}dt$
That extra t in the denominator may be a culprit. Otherwise, it is $I_{n-1}$.
I could repeat as before, but sorry to say I got hung up.
Thanks for your input and help.