How does one get the Bernoulli numbers via the generating function? Here is the definition:

Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$

I've tried to naively expand $\frac{x}{e^x-1}$ around $x_0=0$ and didn't quite understand what should be done with all the $(e^x-1)^{k+1}$ in the denominators of $f^{(k)}(0)$?
How does one get to $$1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \ ...$$ directly from the definition (without rearranging the whole thing into the recursive formula)?
Thanks a lot!

Expand $e^{x}$ around $x=0$ and then reexpand 
\begin{align}\frac{1}{1+\frac{x}{2!}+\frac{x^2}{3!}+\ldots}=1-\left(\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\ldots\right)+\left(\frac{x}{2!}+\frac{x^2}{3!}+\ldots\right)^2-\left(\frac{x}{2!}+\ldots\right)^3+\ldots=\\
=1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\ldots\right)+\frac{x^2}{4}\left(1+\frac{x}{3}+\ldots\right)^2-\frac{x^3}{8}\left(1+\ldots\right)^3+\ldots=\\
=1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}\right)+\frac{x^2}{4}\left(1+\frac{2x}{3}\right)-\frac{x^3}{8}+O\left(x^4\right)=\\
=1-\frac{x}{2}+\left(-\frac{1}{6}+\frac14\right)x^2+\left(-\frac{1}{24}+\frac{1}{6}-\frac{1}{8}\right)x^3+O\left(x^4\right)=\\
=1-\frac{x}{2}+\frac{x^2}{12}+O\left(x^4\right).\end{align}
