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!

Prove by induction that 2^n > n^2 for n ≥ 5. This detailed solution covers the base case, inductive hypothesis, and inductive step, including step-by-step expansions and inequalities. Suitable for advanced high school students (Grades 10-12).

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