The lowest integer which is greater than 1 + 1 10 100 10 100 is

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

Let $x={10}^{100}$

$\Rightarrow P={\left(1+\frac{1}{x}\right)}^x$

$\Rightarrow P=1+\left(x\right)\left(\frac{1}{x}\right)+\frac{\left(x\right)\left(x-1\right)}{\lfloor 2}·\frac{1}{x^2}+\frac{\left(x\right)\left(x-1\right)\left(x-2\right)}{\frac{3}{3}}·\frac{1}{x^3}+\dots$

(upto ${10}^{100}+1$ terms)

$\Rightarrow P=1+1+\left(\frac{1}{\lfloor 2}-\frac{1}{⌊2x}\right)+\left(\frac{1}{\lfloor 3}-\dots \right)+\dots$ so on

$\Rightarrow P=2+$(Positive value less than $\frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\dots$ )

Also $e=1+\frac{1}{\lfloor 1}+\frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\dots$

$\Rightarrow \frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\dots =e-2$

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

$\Rightarrow P\in \left(2,3\right)$

$\Rightarrow$ Least integer value of $P$ is $3.$