The problem asks to prove by induction that $2^n > n$ for all $n \in \mathbb{N}$ (the set of natural numbers).

Step-by-step proof using induction:

Base Case (n = 1):

We first check if the inequality holds for $n = 1$. $2^1 = 2 \quad \text{and} \quad n = 1.$ Clearly, $2 > 1$, so the base case is true.

Induction Hypothesis:

Assume that for some $n = k$, the inequality holds, i.e., $2^k > k.$ This is the induction hypothesis.

Induction Step:

We need to prove that if the inequality holds for $n = k$, then it also holds for $n = k + 1$. That is, we need to show that: $2^{k+1} > k+1.$

Starting with the left-hand side: $2^{k+1} = 2 \cdot 2^k.$ By the induction hypothesis, we know that $2^k > k$, so: $2^{k+1} = 2 \cdot 2^k > 2 \cdot k.$ Thus, we now need to show that $2k \geq k + 1$. Simplifying this inequality: $2k \geq k + 1 \quad \Longleftrightarrow \quad k \geq 1.$ This is true for all $k \geq 1$.

Therefore, the induction step holds, and by the principle of mathematical induction, we have proven that $2^n > n$ for all $n \in \mathbb{N}$.

Would you like further details or clarifications? Here are some related questions:

1. What is the principle of mathematical induction and why is it useful?
2. Can you apply this induction method to other exponential functions, like $3^n > n^2$?
3. What happens if we test the inequality $2^n \geq n$ for $n = 0$?
4. How does the inequality $2^n > n$ behave as $n$ becomes very large?
5. How do we modify the proof if $n \in \mathbb{Z^+}$ (positive integers)?

Tip: Inductive proofs often work well when the statement to be proved involves recursive or iterative structures.