We can simplify the given expression:

\[
\frac{(n+2)!}{(n+1)(n+2)}
\]

### Step-by-step solution:
1. **Expand the factorial** \( (n+2)! \):
   \[
   (n+2)! = (n+2)(n+1)n!
   \]
   
   So, we have:
   \[
   \frac{(n+2)(n+1)n!}{(n+1)(n+2)}
   \]

2. **Cancel out common terms**:
   The terms \( (n+2) \) and \( (n+1) \) appear in both the numerator and the denominator, so they cancel out:
   \[
   \frac{(n+2)(n+1)n!}{(n+1)(n+2)} = n!
   \]

### Final simplified result:
\[
n!
\]

Let me know if you'd like more details or if you have any questions! 

Here are some related questions you might find interesting:
1. How do you compute factorials of large numbers efficiently?
2. What are some real-life applications of factorials in mathematics?
3. How does factorial growth compare to exponential growth?
4. What is the Gamma function, and how is it related to factorials?
5. Can you prove that \( (n+2)! = (n+2)(n+1)n! \) using mathematical induction?

**Tip:** Factorials grow very quickly, and their values become extremely large even for moderately sized integers.

Simplify the factorial expression (n+2)! / (n+1)(n+2) step by step. This problem demonstrates the cancellation of common terms and the expansion of factorials, suitable for Grades 10-12 algebra students.

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