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.