Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF into a series and have a look at the coefficients. Using $e^{f(x)}=1+f(x)+\frac{f(x)^2}{2!}+\frac{f(x)^3}{3!}+\ldots$ I get
\begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{x^2}-2e^x+1}{2!}+\frac{e^{x^3}-3e^{x^2}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{x^2}}{2!}-e^x+\frac{1}{2!}+\frac{e^{x^3}}{3!}-\frac{e^{x^2}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{x^3}}{3!}-\frac{1}{2!}+\frac{1}{3!}\\
&=&\frac{1}{2!}e^x+\frac{1}{3!}e^{x^3}-\frac{1}{3}\\
&=&\frac{1}{2!}\left( 1+x^2+\frac{x^4}{2!}+\frac{x^8}{3!} \right)+\frac{1}{3!}\left( 1+x^3+\frac{x^6}{2!}+\frac{x^9}{3!}\right)\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^8}{24}+\frac{1}{6}+\frac{x^3}{3!}+\frac{x^6}{12}+\frac{x^9}{36}\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^3}{6}+\ldots
\end{eqnarray*}
I think I can stop here, because the coefficient in front of $\frac{x^3}{3!}$ is not $15$.
Perhaps someone can help me out and give a hint to find my mistake?

It's probably easiest to expand the exponential in the exponent first, since that will lead to a finite number of terms to be evaluated:
$$\begin{align}e^{(e^x-1)} &= \exp\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right) +\frac{1}{2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)^2
+\frac{1}{6}\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)^3+O(x^4)\\
&=1+\left(x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\right)+\frac{1}{2}\left(x^2+2(x)\left(\frac{x^2}{2}\right)+O(x^4)\right)+\frac{1}{6}\left(x^3+O(x^4)\right)\\
&=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^3}{6}+O(x^4)\\
&=1+x+x^2+\frac{5x^3}{6}+O(x^4)
\end{align}$$
From which the coefficients can be read off straightforwardly.
