We are tasked with simplifying the expression \(\frac{(n+1)!}{n(n-1)!}\) for \(n > 1\).

### Step-by-step Simplification:

1. Start by expanding \((n+1)!\):
   \[
   (n+1)! = (n+1) \cdot n!
   \]

2. Substitute this into the original expression:
   \[
   \frac{(n+1)!}{n(n-1)!} = \frac{(n+1) \cdot n!}{n(n-1)!}
   \]

3. Notice that \(n! = n \cdot (n-1)!\), so replace \(n!\) with \(n \cdot (n-1)!\) in the numerator:
   \[
   \frac{(n+1) \cdot n \cdot (n-1)!}{n(n-1)!}
   \]

4. Cancel out \(n \cdot (n-1)!\) from both the numerator and denominator:
   \[
   \frac{(n+1)}{1} = n+1
   \]

### Final Result:
The simplified expression is:
\[
n+1
\]

Let me know if you need any clarifications or further details!

### Related Questions:
1. How does the factorial operation change when you shift the variable by a constant (e.g., \( (n+k)! \))?
2. What is the significance of the factorial function in combinatorics?
3. How would the simplification change if the expression involved double factorials?
4. Can you apply similar simplification techniques to expressions involving binomial coefficients?
5. How would the result differ if \(n\) were a non-integer, such as in the Gamma function?

### Tip:
When working with factorial expressions, look for common terms to cancel out early on—this often simplifies the process significantly.

Simplify the factorial expression (n+1)!/n(n−1)! for n > 1 with a detailed step-by-step solution. This problem covers factorial properties and algebraic simplification. Suitable for high school students in Grades 9-12.

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