The coefficient of $$x^n$$ in the expansion of $${e^{{e^x}}}$$ is

Solution :
Let $$e^x = z,$$  then
$$\eqalign{ & {e^{{e^x}}} = {e^z} = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{k!}}} = \sum\limits_{k = 0}^\infty {\frac{{{{\left( {{e^x}} \right)}^k}}}{{k!}}} = \sum\limits_{k = 0}^\infty {\frac{{{e^{kx}}}}{{k!}}} \cr & = \left( {1 + \frac{{{e^x}}}{{1!}} + \frac{{{e^{2x}}}}{{2!}} + \frac{{{e^{3x}}}}{{3!}} + .....\,{\text{to }}\infty } \right) \cr & = 1 + \frac{1}{{1!}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} } \right) + \frac{1}{{2!}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{{\left( {2x} \right)}^n}}}{{n!}}} } \right) + \frac{1}{{3!}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{{\left( {3x} \right)}^n}}}{{n!}}} } \right) + {\text{to}}.....\infty \cr & \therefore {\text{Coefficient of }}{x^n}{\text{ in }}{e^{{e^x}}} \cr & = \frac{1}{{1!}}\left( {\frac{1}{{n!}}} \right) + \frac{1}{{2!}}\left( {\frac{{{2^n}}}{{n!}}} \right) + \frac{1}{{3!}}\left( {\frac{{{3^n}}}{{n!}}} \right) + .....\,{\text{to }}\infty \cr & = \frac{1}{{n!}}\left( {\frac{1}{{1!}} + \frac{{{2^n}}}{{2!}} + \frac{{{3^n}}}{{3!}} + .....\,{\text{to }}\infty } \right) \cr} $$