(related to Proposition: Inequality between Square Numbers and Powers of $2$)

We provide a proof by induction for all natural numbers $n\in\mathbb N.$

- Base case: $n=0$ $$0^2 < 2^0=1.$$
- Induction step $n\to n+1$
- Assume, $$n^2 < 2 ^n$$ is correct for an $n\ge 0.$
- Then $$\begin{align}(n+1)^2&=n^2+2n+1\nonumber\\ & \le n^2+n^2\nonumber\\ &=2\cdot n^2.\nonumber\\ & < 2\cdot 2^n\quad\text{(by assumption)}\nonumber\\ &=2^{n+1}\nonumber \end{align}$$

- Thus, the sum of consecutive odd numbers from $1$ to $n$ equals the square of $n.$∎