To prove the inequality $n! > 2^n$ for $n \geq 4$ by mathematical induction, follow these steps:

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Step 1: Base Case

For $n = 4$: $4! = 4 \times 3 \times 2 \times 1 = 24$ $2^4 = 16$ Clearly, $24 > 16$. So, the base case holds.

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Step 2: Inductive Hypothesis

Assume that the statement $k! > 2^k$ is true for some $k \geq 4$, i.e., $k! > 2^k$ We will prove that $(k+1)! > 2^{k+1}$.

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Step 3: Inductive Step

From the definition of factorial, we know: $(k+1)! = (k+1) \cdot k!$ Using the inductive hypothesis $k! > 2^k$, we substitute: $(k+1)! = (k+1) \cdot k! > (k+1) \cdot 2^k$ Thus, it suffices to prove: $(k+1) \cdot 2^k > 2^{k+1}$ Simplify the right-hand side: $(k+1) \cdot 2^k > 2 \cdot 2^k$ Divide through by $2^k$ (valid since $2^k > 0$): $k+1 > 2$ This is true for $k \geq 4$.

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Conclusion

By the principle of mathematical induction, the inequality $n! > 2^n$ holds for all $n \geq 4$.

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Would you like me to elaborate further on any step or provide examples for deeper clarity? Here are some related questions you might explore:

1. How does induction compare to other proof techniques like contradiction?
2. What are examples of other factorial inequalities?
3. Why is $n \geq 4$ the starting point for this proof?
4. Can you visualize factorial and exponential growth rates graphically?
5. How does Stirling's approximation support this inequality for large $n$?

Tip: When proving inequalities, carefully assess the starting point of validity, as it often provides insight into the inequality's behavior!