The lowest integer which is greater than $$\left(1 + \frac{1}{10^{100}}\right)^{10^{100}}$$ is

We have to find the smallest integer which is strictly greater than the quantity $$\left(1+\frac{1}{10^{100}}\right)^{10^{100}}.$$

First recall the well-known limit definition of the constant e:

$$e=\lim_{m\to\infty}\left(1+\frac1m\right)^m.$$

This result also gives two useful inequalities that hold for every positive integer $$m$$:

$$\left(1+\frac1m\right)^m<e<\left(1+\frac1m\right)^{m+1}.$$

We shall apply these inequalities with the choice

$$m=10^{100}.$$

Substituting $$m=10^{100}$$ into the left-hand member of the double inequality, we get

$$\left(1+\frac1{10^{100}}\right)^{10^{100}}<e.$$

It is a standard numerical fact that

$$e\approx2.718\,281\,8\ldots$$

Therefore

$$\left(1+\frac1{10^{100}}\right)^{10^{100}} < 2.718\,281\,8\ldots$$

So the expression is certainly less than $$3.$$

Next we show that the expression is greater than $$2.$$ To do this, we expand it by the binomial theorem. The binomial theorem states that for any positive integer $$n$$ and any real number $$x,$$

$$\left(1+x\right)^n = 1 + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \cdots.$$

Here $$n=10^{100}$$ and $$x=\dfrac1{10^{100}},$$ so

$$\left(1+\frac1{10^{100}}\right)^{10^{100}} = 1 + \binom{10^{100}}1\frac1{10^{100}} + \binom{10^{100}}2\frac1{10^{200}} + \binom{10^{100}}3\frac1{10^{300}} + \cdots.$$

We evaluate the first two terms explicitly:

$$\binom{10^{100}}1\frac1{10^{100}} = 10^{100}\cdot\frac1{10^{100}} = 1,$$

$$\binom{10^{100}}2\frac1{10^{200}} =\frac{10^{100}(10^{100}-1)}{2}\cdot\frac1{10^{200}} =\frac{10^{100}-1}{2\cdot10^{100}} =\frac12-\frac1{2\cdot10^{100}}.$$

All the remaining terms are positive, so we already have

$$\left(1+\frac1{10^{100}}\right)^{10^{100}} > 1 + 1 = 2.$$

Putting the two inequalities together we obtain

$$2 < \left(1+\frac1{10^{100}}\right)^{10^{100}} < 3.$$

An integer which is strictly greater than the value must therefore be at least $$3,$$ and because the expression is itself less than $$3,$$ the integer $$3$$ is indeed the smallest such integer.

Hence, the correct answer is Option A.