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The problem asks us to find the \( n \)-th term of the sequence \( \{a_n\} \), which is defined as:

\[
a_n = \left[\left(\frac{2}{1}\right)^1 \left(\frac{3}{2}\right)^2 \left(\frac{4}{3}\right)^3 \cdots \left(\frac{n+1}{n}\right)^n\right]^{\frac{1}{n}}.
\]

Let's simplify the expression step-by-step.

### Step 1: Simplify the Product Inside the Bracket

The general term inside the product is \( \left(\frac{k+1}{k}\right)^k \) for \( k \) ranging from 1 to \( n \). So, the entire expression can be rewritten as:

\[
\prod_{k=1}^{n} \left(\frac{k+1}{k}\right)^k.
\]

### Step 2: Rewrite the Product as an Exponential

We can rewrite this as:

\[
\prod_{k=1}^{n} \left(\frac{k+1}{k}\right)^k = \exp\left(\sum_{k=1}^{n} k \ln\left(\frac{k+1}{k}\right)\right).
\]

### Step 3: Simplify the Logarithmic Term

Simplify \( \ln\left(\frac{k+1}{k}\right) \):

\[
\ln\left(\frac{k+1}{k}\right) = \ln(k+1) - \ln(k).
\]

Therefore, the sum becomes:

\[
\sum_{k=1}^{n} k \left(\ln(k+1) - \ln(k)\right).
\]

This expression telescopes when expanded.

### Step 4: Telescoping Series

When expanded, the series will cancel out most terms, resulting in a simplified form. The final simplified result is dominated by the term:

\[
\ln(n+1)^n.
\]

### Step 5: Applying the \( \frac{1}{n} \) Power

The expression inside the original brackets simplifies to \( (n+1)^n \), so applying the \( \frac{1}{n} \) exponent gives:

\[
a_n = (n+1).
\]

### Conclusion

The \( n \)-th term of the sequence \( \{a_n\} \) is:

\[
\boxed{a_n = n+1.}
\]

Would you like further details or have any questions? Here are some related questions:

1. What are telescoping series, and how do they simplify sequences?
2. How do we handle products with exponential terms in sequences?
3. Can we generalize this method to find limits of sequences?
4. What are other ways to evaluate complex sequences?
5. How does this sequence behave as \( n \) approaches infinity?

**Tip:** Simplifying sequences often involves recognizing patterns, especially telescoping terms, which can greatly reduce complexity.

If the n-th term of the sequence {a_n} is given by 
\[
a_n = \left[\left(\frac{2}{1}\right)^1 \left(\frac{3}{2}\right)^2 \left(\frac{4}{3}\right)^3 \cdots \left(\frac{n+1}{n}\right)^n\right]^{\frac{1}{n}}.
\]
Find the n-th term of the sequence.

Learn how to find the n-th term of a sequence defined by a complex product involving exponents and logarithms. This detailed solution simplifies the sequence step-by-step using telescoping series, making it suitable for undergraduate mathematics students.

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