computing the iterated exp(x)-1

[Daniel]

\( f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots \) is the general solution

Which formula for the coefficients is this exactly?

I use what I call Schroeder summations http://tetration.org/Combinatorics/Schro...index.html for all of my iterated calculations. If the iterated function can be differentiated then Schroeder summations can be used, it doesn't matter what type of fixed point is involved.