Taylor series for exp(exp(x)) using just the power series for exp(x) I'm trying to figure out how to calculate the power series for exp(exp(x)) using exp(x) and then to write down the first few terms. I have the answer for the terms but I don't know how they arrived at it 
Thanks in advance
For the first few terms we have: exp(exp(x)) = e(1 + x + x^2 + . . .)
 A: If you want to compute the Taylor series of $e^{e^x}$ up to order $n$, begin with $e^{e^x-1}$. Then$$e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots\tag1$$Now, you forget the terms that come after $\frac{x^n}{n!}$ and, in $(1)$, you replace each $x$ by the Taylor series of $e^x-1$ up to order $n$:\begin{multline}1+\left(x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)+\frac{\left(x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)^2}{2!}+\cdots+\frac{\left(x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}\right)^n}{n!}\end{multline}and you expand this, but then you delete the monomials whose order is greater than $n$:$$1+x+x^2+\frac{5 x^3}{6}+\frac{5 x^4}{8}+\frac{13x^5}{30}+\frac{203 x^6}{720}+\frac{877x^7}{5040}+\cdots$$Finally, you multiply everything by $e$, and you're done.
A: Taking derivatives directly, we have
$$f^{(n)}(x)=\sum_{k=0}^n{n\brace k}\exp(e^x+kx)$$
where $f(x)=\exp(e^x)$ and $n\brace k$ are Stirling numbers of the second kind. From this, we have
$$f^{(n)}(0)=e\sum_{k=0}^n{n\brace k}=eB_n$$
where $B_n$ are the Bell numbers. Thus, we have,

$$\exp(e^x)=e\sum_{n=0}^\infty\frac{B_n}{n!}x^n$$

A: When we put $\cdots$ we'll be ignoring all terms involving $x^n$ for $n \geq 3$. We expand like so:
\begin{align*}
\exp(\exp(x)) &= \exp\left(1 + x + \frac{1}{2}x^2 + O(x^3)\right) \\
&= e \times \exp\left(x + \frac{1}{2} x^2 + O(x^3)\right) \\
&= e\left(1 + \left(x + \frac{1}{2}x^2 + O(x^3)\right) + \frac{1}{2}\left(x + \frac{1}{2}x^2 + O(x^3)\right)^2 + O(x^3) \right) \\
&=e\left(1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^2 + O(x^3) \right) \\
&=e(1 + x + x^2 + O(x^3))
\end{align*}
Normally we'd need to worry about convergence of the series, but in this case the power series for the exponential function has infinite radius of convergence so everything converges.
A: We can use the power series of exp(x) and substitute it into the inner portion of exp(exp(x)), as in:
$$\exp\left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)$$
Then an exponent property can be used, in which, as an example, some $e^{(a+b+c)}$ is just $e^a e^b e^c$.
$$\prod_{n=0}^{\infty} \exp\left(\frac{x^n}{n!}\right)$$ 
Then take the power series of $\exp\left(\frac{x^n}{n!}\right)$ and insert it inside the product:
$$\prod_{n=0}^{\infty} \sum_{p=0}^{\infty} \frac{1}{p!}\left(\frac{x^n}{n!}\right)^p$$
Since you wanted the first few terms, I guess you could let the upper bounds be finite. The following is probably better:
As an example we can write up to the $x^2$ term:
$$\prod_{n=0}^{\infty} \sum_{p=0}^{\infty} \frac{1}{p!}\left(\frac{x^n}{n!}\right)^p$$
$$e\left[1+x+\frac{x^2}{2}+\mathcal{O}(x^3)\right]\left[1+\frac{x^2}{2}+\mathcal{O}(x^3)\right]$$
$$= e+ex^1+ex^2+\mathcal{O}(x^3)$$
