The question asks me to find an exact expression for the n-th derivative of the function $f(x) = e^{e^{x}}$.
I noticed a pattern and claimed that the derivative could be expressed as $$\frac{d^n}{dx^n}[e^{e^{x}}] = S_{1,n}e^xe^{e^{x}} + S_{2,n} e^{2x}e^{e^{x}}+...+S_{n-1,n}e^{(n-1)x}e^{e^{x}}+S_{n,n}e^{nx}e^{e^{x}}$$ where $S_{n,k}$ is the Stirling number of the second kind.
Now I have tried to prove that this is the case by using induction and have easily shown that the base case is true. I am having problem with trying to show the inductive step.
I prefer and will use the notation ${n\brace k}$ for the Stirling numbers of the second kind. Here’s the induction step.
Suppose that
$$\frac{d^n}{dx^n}\left(e^{e^x}\right)=\sum_{k=1}^n{n\brace k}e^{kx}e^{e^x}\;;$$
then
$$\begin{align*} \frac{d^{n+1}}{dx^{n+1}}\left(e^{e^x}\right)&=\sum_{k=1}^n{n\brace k}\left(e^{kx}\cdot e^xe^{e^x}+ke^{kx}e^{e^x}\right)\\ &=\sum_{k=1}^n{n\brace k}\left(e^{k+1}x+ke^{kx}\right)e^{e^x}\\ &=\sum_{k=2}^{n+1}{n\brace{k-1}}e^{kx}e^{e^x}+\sum_{k=1}^nk{n\brace k}e^{kx}e^{e^x}\\ &\overset{(1)}=\sum_{k=1}^{n+1}\left({n\brace{k-1}}+k{n\brace k}\right)e^{kx}e^{e^x}\\ &\overset{(2)}=\sum_{k=1}^{n+1}{{n+1}\brace k}e^{kx}e^{e^x}\;. \end{align*}$$
$(1)$ holds because ${n\brace 0}=0={n\brace{n+1}}$, and $(2)$ uses a standard recurrence satisfied by the Stirling numbers of the second kind.