1. **Problem statement:** Prove by mathematical induction that $n < 2^n$ for all $n \in \mathbb{N}$. 2. **Base case:** For $n=1$, check if $1 < 2^1$. $$1 < 2$$ This is true. 3. **Inductive hypothesis:** Assume for some $k \in \mathbb{N}$, the statement holds: $$k < 2^k$$ 4. **Inductive step:** Prove for $k+1$: $$k+1 < 2^{k+1}$$ Start from the inductive hypothesis: $$k < 2^k$$ Add 1 to both sides: $$k + 1 < 2^k + 1$$ Since $2^k \geq 1$ for all $k \geq 1$, we have: $$2^k + 1 \leq 2^k + 2^k = 2 \cdot 2^k = 2^{k+1}$$ Therefore: $$k + 1 < 2^{k+1}$$ 5. **Conclusion:** By mathematical induction, $n < 2^n$ holds for all $n \in \mathbb{N}$.