We want to show that $2^n > n^3$ for $n \geq 10$. Here's the step-by-step proof:

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Step 1: Verify Base Case ($n = 10$)

Substitute $n = 10$ into $2^n$ and $n^3$: $2^{10} = 1024, \quad n^3 = 10^3 = 1000.$ Clearly: $1024 > 1000.$ So the inequality holds for $n = 10$.

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Step 2: Inductive Step

Use mathematical induction to prove $2^n > n^3$ for all $n \geq 10$.

Inductive Hypothesis

Assume $2^k > k^3$ is true for some $k \geq 10$. That is: $2^k > k^3.$

Inductive Step

We need to show that $2^{k+1} > (k+1)^3$.

Starting from $2^{k+1} = 2 \cdot 2^k$: $2^{k+1} = 2 \cdot 2^k > 2 \cdot k^3, \quad \text{(by the inductive hypothesis)}.$

Now, compare $2 \cdot k^3$ with $(k+1)^3$. Expand $(k+1)^3$: $(k+1)^3 = k^3 + 3k^2 + 3k + 1.$

Thus, we need: $2 \cdot k^3 > k^3 + 3k^2 + 3k + 1.$ Simplify: $k^3 > 3k^2 + 3k + 1.$

Verify for $k \geq 10$

Divide through by $k^2$ (valid since $k > 0$): $k > 3 + \frac{3}{k} + \frac{1}{k^2}.$

For $k \geq 10$, the right-hand side is: $3 + \frac{3}{10} + \frac{1}{100} = 3.31.$

Clearly, $k \geq 10 > 3.31$, so the inequality holds.

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Conclusion

By induction, $2^n > n^3$ for all $n \geq 10$.

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Do you have any questions or want further clarification? Here are some related problems:

1. Show that $3^n > n^4$ for $n \geq 15$.
2. Compare the growth rates of $a^n$ and $n!$ as $n \to \infty$.
3. Prove that $2^n > n^2$ for all $n \geq 5$.
4. Explore the behavior of $n^3 - 2^n$ as $n \to \infty$.
5. Show that $2^n$ grows faster than any polynomial $n^k$ for large $n$.

Tip: Use induction carefully by verifying the base case and comparing terms systematically for the inductive step.