We will prove by mathematical induction that $2^n > n^2$ for all $n \geq 5$.

Step 1: Base Case

We first check the inequality for $n = 5$.

$2^5 = 32 \quad \text{and} \quad 5^2 = 25$ Clearly, $32 > 25$, so the inequality holds for $n = 5$.

Step 2: Inductive Hypothesis

Assume that the inequality holds for some $n = k \geq 5$, i.e.,

$2^k > k^2$ This is our inductive hypothesis.

Step 3: Inductive Step

We need to show that the inequality holds for $n = k+1$, i.e.,

$2^{k+1} > (k+1)^2$

Starting with the left-hand side:

$2^{k+1} = 2 \cdot 2^k$

By the inductive hypothesis, we know $2^k > k^2$, so we substitute this into the inequality:

$2^{k+1} = 2 \cdot 2^k > 2 \cdot k^2$

Thus, we need to prove that:

$2 \cdot k^2 > (k+1)^2$

Expanding the right-hand side:

$(k+1)^2 = k^2 + 2k + 1$

So, the inequality becomes:

$2k^2 > k^2 + 2k + 1$

Subtract $k^2$ from both sides:

$k^2 > 2k + 1$

We now need to show that this inequality holds for $k \geq 5$.

Step 4: Check for $k = 5$

For $k = 5$:

$5^2 = 25 \quad \text{and} \quad 2(5) + 1 = 11$

Clearly, $25 > 11$, so the inequality holds for $k = 5$.

Step 5: Conclusion

Since the inequality $k^2 > 2k + 1$ holds for $k = 5$ and larger values of $k$, we have shown that $2^k > k^2$ implies $2^{k+1} > (k+1)^2$.

Therefore, by the principle of mathematical induction, $2^n > n^2$ for all $n \geq 5$.

Do you have any questions, or would you like further details on a specific step?

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Here are 5 related questions to explore further:

1. Can you use induction to prove $3^n > n^3$ for large $n$?
2. How would the proof change if we start at $n = 1$ instead of $n = 5$?
3. How does the growth rate of exponential functions compare to polynomial functions in general?
4. Can you find counterexamples for small $n$ where $2^n \leq n^2$?
5. How does the binomial expansion help in understanding the inequality $(k+1)^2$?

Tip: Always verify your base case and inductive step carefully—they are essential to the correctness of an induction proof!