Nice puzzle posted by art of mathematics facebook page

Exercise:

Compute, $\int_{0}^{1}{{{e}^{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}}}}{{e}^{{{e}^{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{x}}}}}}}}}}}}\cdots {{e}^{{{e}^{{{e}^{x}}}}}}{{e}^{{{e}^{x}}}}{{e}^{x}}}dx$

Solution: Let’s find the general formula using induction procedure as follows

Clearly, $\int{{{e}^{{{e}^{x}}}}{{e}^{x}}}dx=\int{{{e}^{{{e}^{x}}}}d\left( {{e}^{x}} \right)}={{e}^{{{e}^{x}}}}+c$

Let $u={{e}^{x}}\Leftrightarrow du={{e}^{x}}dx$ , then

$\int{{{e}^{{{e}^{{{e}^{x}}}}}}{{e}^{{{e}^{x}}}}{{e}^{x}}}dx=\int{{{e}^{{{e}^{u}}}}{{e}^{u}}}du=\int{{{e}^{{{e}^{u}}}}d\left( {{e}^{u}} \right)}={{e}^{{{e}^{u}}}}+c={{e}^{{{e}^{{{e}^{x}}}}}}+c$

Thus,$\int{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}}\cdots {{e}^{{{e}^{x}}}}{{e}^{x}}}dx={{e}^{{{e}^{{{}^{{{e}^{{{e}^{x}}}}}}}}}}+c$

Therefore,$\int_{0}^{1}{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}}\cdots {{e}^{{{e}^{x}}}}{{e}^{x}}}dx=\left( {{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}} \right)_{0}^{1}={{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{1}}}}}}}}}}-{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{0}}}}}}}}}}={{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{e}}}}}}}}-{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{e}}}}}}<\infty $