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?

\begin{align}
e^{e^x-1}&= 1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}+\cdots+\frac{(e^x-1)^n}{n!}+ \cdots \\[8pt]
&= \cdots\cdots+\frac{e^{nx}+xe^{(n-1)x}+\binom n2 e^{(n-2)x}+\cdots+ne^x + 1}{n!}+ \cdots\cdots
\end{align}
One of the terms in the expansion of $(e^x-1)^n$ is $e^x$.  When that is expanded, one of the terms will be $x^4/4!$.  No matter how big $n$ is, you don't run out of terms involving $x^4$.  So you don't just get finitely many terms involving $x^4$ and add up their coefficients to see if you get $\dfrac{15x^4}{4!}$.  Rather, you get an infinite series.
The Wikipedia article titled Exponential formula gives a power series expansion of $$e^{f(x)}= \cdots+\frac{b_n x^n}{n!}+\cdots$$ when the power series expansion of $$f(x) = a_1+\frac{a_2 x^2}{2!}+\cdots+\frac{a_n x^n}{n!}+\cdots $$ is known.  Notice this from the article:
$$
b_3 = a_3+3a_2 a_1 + a_1^3
$$
because there is one partition of the set $\{ 1, 2, 3 \}$ that has a single block of size $3$, there are three partitions of $\{ 1, 2, 3 \}$ that split it into a block of size $2$ and a block of size $1$, and there is one partition of $\{ 1, 2, 3 \}$ that splits it into three blocks of size $1$.  Apply that to $4$:
$$
b_4 = a_4 + 4a_3a_1 + 3a_2^2+6a_2a_1^2 + a_1^4
$$
since there is one partition of the set $\{1,2,3,4\}$ that has a single block of size 4; there are four partitions into a block of size $3$ and a block of size $1$; three are three partitions into two blocks of size $2$; there are six partitions into a block of size $2$ and two blocks of size $1$; and there is one partition into four blocks of size $1$.
Wikipedia's article titled Dobinski's formula treats the expansion of each individual Bell number as an infinite series.
There is also Faà di Bruno's formula one the derivatives of composite functions.
