To prove the inequality \(2^n > n\) for all integers \(n \geq 1\) using mathematical induction, we proceed as follows:

### Step 1: Base Case
We check if the inequality holds for \(n = 1\).

\[
2^1 = 2 \quad \text{and} \quad 1 = 1.
\]

Clearly, \(2 > 1\), so the base case holds.

### Step 2: Inductive Hypothesis
Assume that the inequality \(2^k > k\) holds for some arbitrary positive integer \(k\). That is, we assume:

\[
2^k > k.
\]

### Step 3: Inductive Step
We need to show that \(2^{k+1} > k+1\), assuming the inductive hypothesis \(2^k > k\).

Now, consider:

\[
2^{k+1} = 2 \cdot 2^k.
\]

Using the inductive hypothesis, we know that \(2^k > k\). Thus, multiplying both sides of this inequality by 2 gives:

\[
2^{k+1} = 2 \cdot 2^k > 2 \cdot k.
\]

Therefore, we need to show that:

\[
2 \cdot k \geq k + 1.
\]

Simplifying the inequality:

\[
2k \geq k + 1 \quad \Longleftrightarrow \quad k \geq 1.
\]

This is true because \(k \geq 1\) for all positive integers \(k\). Thus, we have shown that:

\[
2^{k+1} > k + 1.
\]

### Conclusion:
By mathematical induction, the inequality \(2^n > n\) holds for all integers \(n \geq 1\).

---

Would you like further clarification on any part of this proof?

Here are five related questions:
1. How do we prove more complex inequalities using induction?
2. Can the same approach be used to prove \(3^n > n^2\)?
3. What is the significance of base cases in induction?
4. How do we choose an appropriate inductive hypothesis for different problems?
5. How does strong induction differ from regular induction?

**Tip:** Always make sure to clearly state the inductive hypothesis and how it is used in the inductive step.

Learn how to prove the inequality 2^n > n for all integers n ≥ 1 using mathematical induction. This problem explores key concepts such as exponential functions, inequalities, and the principle of mathematical induction. Suitable for high school students in Grades 9-12.

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