1. Let $N=\left(1+\frac{1}{10^{100}}\right)^{10^{100}}.$ We need the lowest integer greater than $N$, i.e. $\lceil N\rceil$.

2. This is a standard expression related to the number $e$: $\left(1+\frac{1}{n}\right)^n<e<3 \quad \text{for all positive integers } n.$ Here $n=10^{100}$, so $\left(1+\frac{1}{10^{100}}\right)^{10^{100}}<e.$ Since $e\approx 2.71828<3,$ we get $N<3.$

3. Also, $\left(1+\frac{1}{10^{100}}\right)^{10^{100}}>1+10^{100}\cdot \frac{1}{10^{100}}=2,$ by the binomial expansion (all remaining terms are positive). So, $N>2.$

4. Therefore, $2<N<3.$ Hence the smallest integer greater than $N$ is $3.$

5. Checking options:

• A: $3$ ✅
• B: $4$ ❌
• C: $2$ ❌
• D: $1$ ❌

Therefore, the correct answer is A.