Let $P=\left(1+\frac{1}{10^{100}}\right)^{10^{100}}$
Let $x=10^{100}$
$\Rightarrow P=\left(1+\frac{1}{x}\right)^{x}$
$\Rightarrow P=1+(x)\left(\frac{1}{x}\right)+\frac{(x)(x-1)}{2 !} \cdot \frac{1}{x^{2}}+\frac{(x)(x-1)(x-2)}{3 !} \cdot \frac{1}{x^{3}}+\ldots \ldots \ldots$ (upto $10^{100}+1$ terms )
$\Rightarrow P=1+1+\left(\frac{1}{2 !}-\frac{1}{2 ! x^{2}}\right)+\left(\frac{1}{3 !}-\ldots \ldots\right)+$ so on
$\Rightarrow P=2+\left(\right.$ Positive value less then $\left.\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots .\right)$
Also e $=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$
$\Rightarrow \frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots .=e-2$
$\Rightarrow P=2+($ positive value less then $e-2)$
$\Rightarrow P \in(2,3)$
$\Rightarrow$ least integer value of $P$ is $3$