Strategy for expanding the function $e^{e^x-1}$ into Maclaurin series I am trying to find a way to expand the function $e^{e^x-1}$ into Maclaurin series but I don't know how to. I don't want to use the cumbersome Maclaurin formula of taking successive higher derivative.
Do you know how to nicely expand this function into Taylor series. I know that this is the generating function of Bell numbers.
Wolfram gives the expansion as:
$1-x+x^2+\dfrac{5x^3}{6}+\dfrac{5x^4}{8}...$

The exponent is
\begin{eqnarray*}
e^x-1 = x+\frac{x^2}{2}+ \frac{x^3}{6}+\cdots.
\end{eqnarray*}
Now just plug this in the power series for the exponential function
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = 1&+& \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right) \\
&+& \frac{1}{2} \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right)^2 \\
&+& \frac{1}{6} \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right)^3 \\
&+& \cdots  \\
\end{eqnarray*}
Expand to whatever order you need ...
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = \color{red}{1}+\color{red}{1}x+ \frac{\color{red}{2}x^2}{2}+ \frac{\color{red}{5}x^3}{3!} +\cdots
\end{eqnarray*}
But of course ... as Ethan Bolker says in the comments the Bell numbers can be calculated much more easily using
\begin{eqnarray*}
B_{n+1} = \sum_{k=0}^{n}  \binom{n}{k} B_k.
\end{eqnarray*}
