The inequality \( n! > 2^n \) holds for \( n \geq 4 \). To verify this, let's check a few values manually first and then generalize the result.

1. For \( n = 4 \):
   \[
   4! = 4 \times 3 \times 2 \times 1 = 24
   \]
   \[
   2^4 = 16
   \]
   So, \( 4! = 24 > 16 = 2^4 \).

2. For \( n = 5 \):
   \[
   5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
   \]
   \[
   2^5 = 32
   \]
   So, \( 5! = 120 > 32 = 2^5 \).

3. For \( n = 6 \):
   \[
   6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
   \]
   \[
   2^6 = 64
   \]
   So, \( 6! = 720 > 64 = 2^6 \).

As you can see, starting from \( n = 4 \), the factorial function grows much faster than the exponential function \( 2^n \). Therefore, the inequality \( n! > 2^n \) holds for all \( n \geq 4 \).

### Summary:
For all \( n \geq 4 \), \( n! > 2^n \).

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Do you want more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How can we rigorously prove that \( n! > 2^n \) for all \( n \geq 4 \)?
2. What is the asymptotic behavior of factorial and exponential functions as \( n \to \infty \)?
3. How does Stirling’s approximation help in comparing \( n! \) and \( 2^n \)?
4. Can you find another inequality involving \( n! \) and a different exponential function?
5. How does the factorial growth compare with other common functions like polynomials or logarithms?

**Tip:** Factorial functions grow extremely fast compared to exponential functions, making them useful for analyzing large combinatorial problems.

The query submitted by the user: n! > 2^n for all n ≥ 4

Explore the mathematical inequality n! > 2^n for all n ≥ 4. This problem involves key concepts such as factorials and exponential functions, highlighting the rapid growth of factorials compared to exponentials. Suitable for students in Grades 10-12.

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