Laurent series expansion about z=0 I am to obtain the first 3 terms of the Laurent series about $z = 0$ for:
$$f(z)=1/(e^z-1-z)$$
I know that there is 1 singularity at $z=0$. My denominator is not in a polynomial form so I can convert it into such by taking a Taylor series of the function around z=0 which yields the denominator as:
$$(e^z-1-z) = x^2/2 + x^3/6 + x^4/24 + ...$$
This is the part where I get stuck. I'm supposed to bring that back in and perhaps factor something out of the denominator so the left side is analytic at z=0 and the singularities are shifted to the term on the right side. Any tips on what my next step should be? Many thanks!

Borrowing from that answer I commented, we have
\begin{align}
  \frac{1}{e^z - 1 - z} &= \frac{1}{(1 + z + z^2/2! + \dots) - 1 - z}\\
  &= \frac{1}{z^2/2! + z^3/3! + z^4/4! + \dots}\\
  &= \frac{2}{z^2} \cdot \frac{1}{1 + 2z/3! + 2z^2/4! + \dots}
\end{align}
Now letting $P(z) = 2z/3! + 2z^2/4! + \dots$, we have
\begin{align}
  \frac{1}{e^z - 1 - z} &= \frac{2}{z^2} \cdot (1 - P(z) + P(z)^2 - P(z)^3 + \dots)\\
  &= \frac{2}{z^2} \cdot (1 - 2z/3! + z^2(-2/4! + (2/3!)^2) + \dots)\\
  &= 2z^{-2} - \frac{2z^{-1}}{3} + \frac{1}{18} - \dots
\end{align}
Getting higher order terms starts to get a bit complicated, so I'd suggest using Wolfram Alpha to fill in the rest: http://www.wolframalpha.com/input/?i=laurent+series+of+1%2F(e%5Ez+-+1+-+z)
