[Math] Proof that the sum of the first $n$ odd numbers is $n^2$.

Question

Here is what I have so far:

The $n$th odd number is $2n-1$.

So we prove that $1+3+…+(2n-3)+(2n-1)= n^2$.

Separate the last term and you get: $[1+3+…+(2n-3)]+(2n-1)$

$[1+3+…+(2n-3)]$ is the sum of the first $(n-1)$ odd numbers.

Here is where I get stuck. The textbook says that the sum of the first $(n-1)$ odd numbers is $(n-1)^2$, but why is that the case? It seems like a recursive explanation because we are trying to prove that the sum of the first $n$ odd numbers is $n^2$. Since we have not yet proved that, how can one say with certainty that the sum of the first $(n-1)$ odd numbers is $(n-1)^2$ ?

Answer 1

Here is a proof without words: