So I tried using Leibnitz formula to solve by recurrence, but I can just get to one point and then it's a mess again. Problem is
Let $f(x)=\frac{1}{1+2x+3x^2+\ldots+2005x^{2004}}$. Find $f^{[2005]}(0)$.
What I did is to notice that $f'=-f^2g$, where $g$ is the polynomial, and then tried to expand the result to higher order derivatives. By Leibnitz formula and the last expression, I got
$f^{[2005]}(0)=\displaystyle\sum_{k=0}^{2004}\binom{2004}{k}\left(f^2\right)^{[k]}(0)g^{[2004-k]}(0)=\sum_{k=0}^{2004}\binom{2004}{k}\left(f^2\right)^{[k]}(0)(2004-k)! =\sum_{k=0}^{2004}\frac{2004!}{k!}\left(f^2\right)^{[k]}(0)$
I'd really appreciate some help here, please.
Notice $$\frac{1}{f(x)} =\sum_{k=0}^{2004} (k+1)x^k = \left(\sum_{k=0}^{\infty} - \sum_{k=2005}^{\infty}\right) (k+1)x^k = \frac{1}{(1-x)^2} - 2006 x^{2005} + O(x^{2006}) $$ We have $$f(x) = \frac{(1-x)^2}{1 - 2006 x^{2005} + O(x^{2006})} = (1-x)^2\left( 1 + 2006 x^{2005} + O(x^{2006})\right)\\ = (1-x)^2 + 2006 x^{2005} + O(x^{2006}) $$ This implies $$f^{[2005]}(0) = 2006 \times 2005! = 2006!$$