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 since we already have the base case as n = 0, we can assume k > 0 for our inductive hypothesis.

To prove the inequality \( 2^n > n \) for all integers \( n \geq 0 \) using mathematical induction, we can follow these steps:

### Step 1: Base Case
We check the base case \( n = 0 \):

\[
2^0 = 1 \quad \text{and} \quad 1 > 0
\]

The inequality \( 2^n > n \) holds true for \( n = 0 \).

### Step 2: Inductive Hypothesis
Assume that the inequality \( 2^k > k \) holds for some integer \( k \geq 0 \). This is our **inductive hypothesis**.

### Step 3: Inductive Step
We need to show that if \( 2^k > k \), then \( 2^{k+1} > k + 1 \) as well.

Starting with the left side, we use the property of exponents to express \( 2^{k+1} \):

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

Since we assume \( 2^k > k \), we can multiply both sides of this inequality by 2:

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

Thus,

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

Now, we need to show that \( 2k \geq k + 1 \), which would imply \( 2^{k+1} > k + 1 \).

Rearrange \( 2k \geq k + 1 \) to \( k \geq 1 \). Therefore, for \( k \geq 1 \

Prove using induction that 2^n > n for all integers n ≥ 0.

Understand how to prove the inequality 2^n > n for all integers n ≥ 0 using mathematical induction. This problem covers induction steps, inequality manipulation, and exponent properties. Suitable for high school students in Grades 10-12.

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