computing the iterated exp(x)-1

[jaydfox]

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

Which formula for the coefficients is this exactly?

I get a similar formula.

\(
\begin{eqnarray}
f(z) & = & e^z-1 \\
\\[10pt]

\\
f^{\circ n}(z) & = &
\left(\frac{1}{2^0}\right)\left(n^0\right)z^1\ +\\ & &
\left(\frac{1}{2^1}\right)\left(n^1\right)z^2\ +\\ & &
\left(\frac{1}{2^2}\right)\left(n^2-\frac{n}{3}\right){z^3}\ +\\ & &
\left(\frac{1}{2^3}\right)\left(n^3-\frac{5n^2}{6}+\frac{n}{6}\right)z^4\ +\\ & &
\left(\frac{1}{2^4}\right)\left(n^4-\frac{13n^3}{9}+\frac{2n^2}{3}-\frac{4n}{45}\right)z^5\ +\\ & &
\left(\frac{1}{2^5}\right)\left(n^5-\frac{77n^4}{36}+\frac{89n^3}{54}-\frac{91n^2}{180}+\frac{11n}{270}\right)z^6\ +\\ & &
\left(\frac{1}{2^6}\right)\left(n^6-\frac{29n^5}{10}+\frac{175n^4}{54}-\frac{149n^3}{90}+\frac{91n^2}{270}-\frac{n}{105}\right)z^7\ + \\ & &
\left(\frac{1}{2^7}\right)\left(n^7-\frac{223n^6}{60}+\frac{1501n^5}{270}-\frac{37n^4}{9}+\frac{391n^3}{270}-\frac{43n^2}{252}-\frac{11n}{1890}\right)z^8\ + \\ & &
\left(\frac{1}{2^8}\right)\left(n^8-\frac{481n^7}{105}+\frac{2821n^6}{324}-\frac{13943n^5}{1620}+\frac{725n^4}{162}-\frac{2357n^3}{2268}+\frac{17n^2}{420}+\frac{29n}{5670}\right)z^9\ +\ \cdots
\end{eqnarray}
\)

It's hard to pick out any particular pattern, but I gave it an initial try...

Edit: Removed an extra factorial that I had added to make finding the fractions easier. I forgot to pull the factorial back out.

Edit #2: I removed the second set of factorials because they cancel out. They were useful for seeing the growth rate, but they made the formula look more complicated than it really is.