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?

You probably don't want to use the series expansion for the exponential as you've done, because you will receive contributions to each coefficient from all orders of this expansion, so you'll still have infinite sums left to evaluate.  Instead, you should use the fact that term $n$ in a sequence is given by $F^{(n)}(0)$, where $F(x)$ is the sequence's exponential generating function (and $F^{(n)}$ is the $n$-th derivative of $F$).  If $F(x)=e^{f(x)}$, then its derivatives are given by
$$
F(0)=e^{f(0)} \\
F'(0)=f'(0)e^{f(0)} \\
F''(0)=(f''(0)+f'(0)^2)e^{f(0)} \\
F'''(0)=(f'''(0)+3f''(0)^2 f'(0) + f'(0)^3)e^{f(0)} \\
F^{(4)}(0)=(f^{(4)}(0)+f'''(0)f'(0)+6f'''(0)f''(0)f'(0)+3f''(0)^2+3f''(0)f'(0)^2+f'(0)^4)e^{f(0)}
$$
and so on.  If $f(x)=e^x-1$, then $e^{f(0)}=1$ and $f^{(n)}(0)=1$ for any $n \ge 1$.  Adding up the terms, then, you have $B_0=B_1=1$, $B_2=2$, $B_3=5$, $B_4=15$, and so on.
