Sum of the First $n$ Odd Numbers

Statement

For all $n\in\mathbb{N}$, \[ 1+3+5+\cdots+(2n-1)=n^2. \]

Proof (Induction)

Base case $n=1$: \[ 1 = 1^2. \] Inductive step: assume \[ 1+3+\cdots+(2n-1)=n^2. \] Then \[ 1+3+\cdots+(2n-1)+(2(n+1)-1) = n^2+(2n+1) = (n+1)^2. \] Hence true for all $n\in\mathbb{N}$.