The given expression is:

$\left( 1 + \frac{1}{2} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{4} \right) \dots \left( 1 + \frac{1}{2025} \right)$

This can be simplified as:

$\prod_{n=2}^{2025} \left( 1 + \frac{1}{n} \right)$

Each term $1 + \frac{1}{n}$ simplifies to:

$1 + \frac{1}{n} = \frac{n + 1}{n}$

So, the entire product becomes:

$\prod_{n=2}^{2025} \frac{n+1}{n}$

Now, let's expand a few terms to understand the pattern:

$\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{2026}{2025}$

Notice that this is a telescoping product where most intermediate terms cancel out. Specifically, all terms cancel except for the first denominator $2$ and the last numerator $2026$. So, the product simplifies to:

$\frac{2026}{2} = 1013$

Thus, the value of the given expression is $\boxed{1013}$.

Would you like a more detailed explanation of the telescoping product? Let me know if you have any further questions.

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Here are 5 related questions to expand on the topic:

1. How do telescoping series and products help simplify complex expressions?
2. What is the general method to compute products of terms like $\prod_{n=a}^{b} \left( 1 + \frac{1}{n} \right)$?
3. How can we apply this approach to sums or integrals?
4. Can we generalize the product formula for any sequence of fractions like $\frac{n+1}{n}$?
5. What are some other types of simplifications often used in sequences and series?

Tip: When dealing with telescoping series or products, always look for terms that cancel out to significantly reduce the complexity of the calculation.