The lowest integer which is greater than (1 + 1 10 ^ 100 ) ^ 10 ^...

Consider $P = {\left(1 + \dfrac{1}{{10}^{100}}\right)}^{{10}^{100}},$

Let $x = {10}^{100}$

$⇒ P = {\left(1 + \dfrac{1}{x}\right)}^{x}$

$⇒ P = 1 + \left(x\right)\left(\dfrac{1}{x}\right) + \dfrac{\left(x\right)\left(x - 1\right)}{⌊ 2} \cdot \dfrac{1}{{x}^{2}} + \dfrac{\left(x\right)\left(x - 1\right)\left(x - 2\right)}{\dfrac{3}{3}} \cdot \dfrac{1}{{x}^{3}} + \ldots$

(upto ${10}^{100} + 1$ terms)

⇒ P = 1 + 1 + \left(\dfrac{1}{⌊ 2} - \dfrac{1}{\left⌊2x\right\}\right) + \left(\dfrac{1}{⌊ 3} - \ldots\right) + \ldots so on

$⇒ P = 2 +$(Positive value less than $\dfrac{1}{⌊ 2} + \dfrac{1}{⌊ 3} + \dfrac{1}{⌊ 4} + \ldots$ )

Also $e = 1 + \dfrac{1}{⌊ 1} + \dfrac{1}{⌊ 2} + \dfrac{1}{⌊ 3} + \dfrac{1}{⌊ 4} + \ldots$

$⇒ \dfrac{1}{⌊ 2} + \dfrac{1}{⌊ 3} + \dfrac{1}{⌊ 4} + \ldots = e - 2$

$⇒ P = 2 +$ (Positive value less than $e - 2$)

$⇒ P \in \left(2, 3\right)$

$⇒$ Least integer value of $P$ is $3 .$