Find the coefficient of $x^r$ in the expansion of $e^{e^x}$.

Question

Find the coefficient of $x^r$ in the expansion of $e^{e^x}$.

My Attempt:

$$e^{e^x}= 1+\dfrac {e^x}{1!} + \dfrac {(e^x)^{2}}{2!} + \dfrac {(e^x)^{3}}{3!}+......$$

How to proceed further?

Answer 1

Let $\sum_{n\ge0}\frac{u_n}{n!}x^n=e^{e^x}$, which multiplies by $e^x$ when differentiated, viz. $$\sum_{n\ge0}\frac{u_{n+1}}{n!}x^n=\sum_{n\ge0}\frac{u_n}{n!}x^n\sum_{n\ge0}\frac{1}{n!}x^n$$ so $u_{n+1}=\sum_{k=0}^n\binom{n}{k}u_k$. Combining this with $u_0=e^{e^0}=e$, in terms of Bell numbers $u_n=eB_n$.