Understanding the Core Concepts

This problem leverages two fundamental mathematical concepts:

1. The Binomial Theorem: For any positive integer $n$ and any real numbers $a$ and $b$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \dots + b^n$ A more specific form, often used in calculus and approximations, is for $(1+y)^n$: $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots + y^n$
2. The Mathematical Constant $e$: The constant $e$ (Euler's number) is defined by several equivalent expressions. The most relevant for this problem is the limit definition: $e = \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$ And its infinite series expansion: $e = \sum_{n=0}^\infty \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$

Step-by-Step Solution

1. Simplify the Expression with Substitution

Let the given expression be $P$: $P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$ To make the expression more manageable and to clearly relate it to the definition of $e$, we perform a substitution. Let $x = {10^{100}}$. Reasoning: This substitution simplifies the appearance of the expression and highlights its structure as $\left(1 + \frac{1}{x}\right)^x$, which is directly related to the definition of $e$. Since $10^{100}$ is an extremely large positive integer, $x$ approaches infinity conceptually for this type of problem.

Now, $P$ becomes: $P = {\left( {1 + {1 \over x}} \right)^x}$

2. Apply the Binomial Theorem

Since $x = {10^{100}}$ is a positive integer, we can expand ${\left( {1 + {1 \over x}} \right)^x}$ using the binomial theorem. Reasoning: The binomial theorem provides an exact expansion for expressions raised to a positive integer power. This allows us to analyze the individual terms and compare them to the series expansion of $e$.

Using the binomial expansion $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$, with $y = \frac{1}{x}$ and $n = x$: $P = 1 + (x)\left( {{1 \over x}} \right) + \frac{x(x-1)}{2!}\left( {{1 \over x}} \right)^2 + \frac{x(x-1)(x-2)}{3!}\left( {{1 \over x}} \right)^3 + \dots + \left( {{1 \over x}} \right)^x$ Tip: Note that this expansion will have $x+1$ terms because the exponent is $x$.

3. Simplify the Terms of the Expansion

Now, we simplify each term: $P = 1 + 1 + \frac{x(x-1)}{2!x^2} + \frac{x(x-1)(x-2)}{3!x^3} + \dots$ Let's simplify the fractional coefficients: $P = 1 + 1 + \frac{1}{2!} \frac{x(x-1)}{x^2} + \frac{1}{3!} \frac{x(x-1)(x-2)}{x^3} + \dots$ $P = 1 + 1 + \frac{1}{2!} \left(1 - \frac{1}{x}\right) + \frac{1}{3!} \left(1 - \frac{1}{x}\right)\left(1 - \frac{2}{x}\right) + \dots$ Reasoning: This simplification rearranges the terms into a form that is directly comparable to the infinite series definition of $e$. By dividing each factor by $x$, we reveal terms like $\left(1 - \frac{k}{x}\right)$, which are slightly less than 1.

4. Compare with the Series Expansion of $e$

Recall the infinite series for $e$: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$ Now, let's compare $P$ with $e$ term by term: $P = 1 + \frac{1}{1!} + \frac{1}{2!} \left(1 - \frac{1}{x}\right) + \frac{1}{3!} \left(1 - \frac{1}{x}\right)\left(1 - \frac{2}{x}\right) + \frac{1}{4!} \left(1 - \frac{1}{x}\right)\left(1 - \frac{2}{x}\right)\left(1 - \frac{3}{x}\right) + \dots$ Reasoning: This comparison is crucial for establishing the bounds of $P$. We observe that for $k \ge 1$, the terms $\left(1 - \frac{k}{x}\right)$ are all less than 1, since $x = 10^{100}$ is a very large positive number.

5. Establish Bounds for $P$

From the expansion of $P$:

• The first two terms are $1+1 = 2$.
• For every subsequent term $\frac{1}{k!} \left(1 - \frac{1}{x}\right)\left(1 - \frac{2}{x}\right)\dots\left(1 - \frac{k-1}{x}\right)$, since $x = 10^{100}$ is a finite (though very large) positive number, each factor $\left(1 - \frac{j}{x}\right)$ for $j < k$ will be positive and strictly less than 1.
  • This means $\left(1 - \frac{1}{x}\right) < 1$
  • $\left(1 - \frac{1}{x}\right)\left(1 - \frac{2}{x}\right) < 1$
  • And so on.

Therefore, for $k \ge 2$: $\frac{1}{k!} \left(1 - \frac{1}{x}\right)\left(1 - \frac{2}{x}\right)\dots\left(1 - \frac{k-1}{x}\right) < \frac{1}{k!}$ Reasoning: Because each multiplying factor $\left(1 - \frac{j}{x}\right)$ is less than 1, the entire term in the expansion of $P$ is strictly smaller than the corresponding term in the infinite series for $e$ (starting from $k=2$).

This leads to the inequality: $P < 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ $P < e$ Also, it's clear that $P > 2$ since the first two terms sum to 2, and all subsequent terms are positive.

Combining these, we get: $2 < P < e$

6. Determine the Lowest Integer Greater than $P$

We know that $e \approx 2.71828$. Since $2 < P < e$, we have $2 < P < 2.71828$. Reasoning: With the numerical bounds established, we can definitively locate $P$ between two integers.

The lowest integer which is strictly greater than $P$ must be $3$.

Summary and Key Takeaway

The value of the expression ${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$ is slightly less than $e$ (approximately $2.718$). By using the binomial expansion and comparing it term-by-term with the infinite series definition of $e$, we established that $2 < P < e$. Therefore, the lowest integer strictly greater than $P$ is $3$. This problem elegantly demonstrates the relationship between the binomial theorem and the fundamental constant $e$.