The value of $$\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt 2 } } } } = ?$$

Solution(By Examveda Team)

$$\eqalign{ & \sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt 2 } } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{{2.2}^{\frac{1}{2}}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{2^{\left( {1 + \frac{1}{2}} \right)}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{2^{\frac{3}{2}}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2.{{\left( {{2^{\frac{3}{2}}}} \right)}^{\frac{1}{2}}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{{2.2}^{\left( {\frac{3}{2} \times \frac{1}{2}} \right)}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{{2.2}^{\frac{3}{4}}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{2^{\left( {1 + \frac{3}{4}} \right)}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{2^{\frac{7}{4}}}} } } \cr & = \sqrt {2\sqrt {2{{\left( {{2^{\frac{7}{4}}}} \right)}^{\frac{1}{2}}}} } \cr & = \sqrt {2\sqrt {{{2.2}^{\left( {\frac{7}{4} \times \frac{1}{2}} \right)}}} } \cr & = \sqrt {2\sqrt {{{2.2}^{\frac{7}{8}}}} } \cr & = \sqrt {2\sqrt {{2^{\left( {1 + \frac{7}{8}} \right)}}} } \cr & = \sqrt {2\sqrt {{2^{\frac{{15}}{8}}}} } \cr & = \sqrt {2{{\left( {{2^{\frac{{15}}{8}}}} \right)}^{\frac{1}{2}}}} \cr & = \sqrt {{{2.2}^{\frac{{15}}{{16}}}}} \cr & = \sqrt {{2^{\left( {1 + \frac{{15}}{{16}}} \right)}}} \cr & = \sqrt {{2^{\frac{{31}}{{16}}}}} \cr & = {\left( {{2^{\frac{{31}}{{16}}}}} \right)^{\frac{1}{2}}} \cr & = {2^{\left( {\frac{{31}}{{16}} \times \frac{1}{2}} \right)}} \cr & = {2^{\frac{{31}}{{32}}}} \cr} $$