We are tasked with simplifying the following product:

$\left( 1 - \frac{1}{2^2} \right) \left( 1 - \frac{1}{3^2} \right) \left( 1 - \frac{1}{4^2} \right) \dots \left( 1 - \frac{1}{100^2} \right)$

Each term in this product takes the form $\left( 1 - \frac{1}{n^2} \right)$, which can be simplified using the difference of squares:

$1 - \frac{1}{n^2} = \frac{n^2 - 1}{n^2} = \frac{(n-1)(n+1)}{n^2}$

Therefore, the product becomes:

$\prod_{n=2}^{100} \left( 1 - \frac{1}{n^2} \right) = \prod_{n=2}^{100} \frac{(n-1)(n+1)}{n^2}$

This product telescopes, meaning that many terms cancel out. Specifically, the numerator and denominator of consecutive terms cancel with each other. Let's write out a few terms to see the cancellation pattern:

$\frac{1 \cdot 3}{2^2} \times \frac{2 \cdot 4}{3^2} \times \frac{3 \cdot 5}{4^2} \times \dots \times \frac{99 \cdot 101}{100^2}$

Almost everything cancels, and what remains is:

$\frac{1 \cdot 101}{100^2} = \frac{101}{10000}$

Thus, the product simplifies to:

$\frac{101}{10000}$

This means that $p = 101$ and $q = 10000$, so:

$\frac{p}{q} = \frac{101}{10000}$

Would you like more details on the cancellation process or have any questions?

Here are some related questions for further exploration:

1. How does the telescoping product work in general?
2. Can you apply this technique to other types of sequences?
3. How can you simplify other complex products involving similar terms?
4. What is the significance of telescoping in calculus?
5. How would the answer change if the upper bound were a different number (e.g., 50 or 200)?

Tip: Telescoping products or sums are powerful tools for simplifying complex expressions, as many intermediate terms cancel out.