The positive integer just greater than ${(1+0.0001)}^{10000}$ , is

For any $n\in N$, we have 

$\begin{array}{l}{\left(1+\frac{1}{n}\right)}^n=1+n\cdot \frac{1}{n}+\frac{n(n-1)}{2!}{\left(\frac{1}{n}\right)}^2+\frac{n(n-1)(n-2)}{3!}{\left(\frac{1}{n}\right)}^3+\dots \\ \Rightarrow {\left(1+\frac{1}{n}\right)}^n=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\dots \\ \Rightarrow {\left(1+\frac{1}{n}\right)}^n=2+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\dots \end{array}$

$\Rightarrow {\left(1+\frac{1}{n}\right)}^n=2+$a positive real number

$\Rightarrow {\left(1+\frac{1}{n}\right)}^n>2$ for al $n\in N$

We know that  

$\underset{n\to \mathrm{\infty }}{lim} {\left(1+\frac{1}{n}\right)}^n=e\approx 2.71$

$\Rightarrow 2<{\left(1+\frac{1}{n}\right)}^n<3$ for all $n\in N$

$\Rightarrow 2<(1+0.0001{)}^{10000}<3$ $[\text{ Putting }n=10000]$

Hence, the greatest integer just greater than $(1+0.0001{)}^{10000}$ is $3$.