Claim: the sum of the first $n$ odd numbers $d_1+\cdots +d_n=n^2$.

We will proceed through induction on $n$.

Base Case: let $n=1$. Then the "sum" of this first odd number is just $d_1=2-1=1$, which is equal to $1^2$.

Induction Hypothesis: Assume that the sum of the first $n-1$ odd numbers is

$d_1+\cdots +d_{n-1}=(n-1{)}^2$.

Inductive Step: Consider the sum of the first $n$ odd numbers $d_1+\cdots +d_n$.

We can split this sum into two parts (using parentheses):

$(d_1+\cdots +d_{n-1})+d_n$.

The term in the parentheses is exactly the sum of the first $n-1$ odd numbers. Note that this is $(n-1{)}^2$ by our Induction Hypothesis. So we can write the sum as

$(n-1{)}^2+d_n$.

Now note that $d_n=2n-1$ by definition. Moreover, we can expand $(n-1{)}^2=n^2-2n+1$. So our expression is now

$(n^2-2n+1)+(2n-1)=n^2$.

Thus the sum of the first $n$ odd numbers is $n^2$.