The value of ${\log }_2. {\log }_3...{\log }_{100} {100}^{{99}^{{98}^{.{ }^{.{ }^{.{ }^{2^1}}}}}} \mathrm{is}$

${\log }_2.{\log }_3...{\log }_{99} {\log }_{100} {100}^{{99}^{{98}^{.{ }^{.{ }^{.{ }^{2^1}}}}}}$

$={\log }_2 {\log }_3 ...{\log }_{99} {99}^{{98}^{.{ }^{.{ }^{.{ }^{2^1}}}}} [{\log }_{100} 100=1]$

$={\log }_2 {\log }_3 ...{\log }_{98} {98}^{{97}^{.{ }^{.{ }^{2^1}}}}$

$={\log }_2\left({\log }_3\left(3^2\right)\right)$

$={\log }_2\left(2\right)$

$=1$