Proof by Induction: Solving $1+3+5+\cdots+(2n-1)$

The question asks to verify that each equation is true for every positive integer n.

The question is as follows:

$$1+ 3 + 5 + \cdots + (2n - 1) = n^2$$

I have solved the base step which is where $n = 1$.

However now once I proceed to the inductive step, I get a little lost on where to go next:

Assuming that k is true (k = n), solve for k+1:

(2k - 1) + (2(k+1) - 1)
(2k - 1) + (2k+2 - 1)
(2k - 1) + (2k + 1)

This is where I am stuck. Do I factor these further to obtain a polynomial of some sort? Or am I missing something?

Assume true for $k$. Then consider the case $k+1$, you got $$1+3+\cdots+(2k-1)+(2(k+1)-1)$$ which is equal by inductive hypothesis $$k^2+(2k+1)=(k+1)^2$$ and this closes the induction.

You have $$ 1+3+5+7+\cdots+(2k-1)\quad + \quad \Big( 2(k+1) - 1 \Big). $$ This is equal to $$ k^2 \quad + \quad \Big( 2(k+1) - 1\Big). $$ Simplify: $$ k^2 + 2k + 2 - 1 $$ $$ = k^2 + 2k + 1 $$ $$ = (k+1)^2. $$

Even though there is an accepted answer, I feel compelled to give a "geometric proof," and by that, I mean a picture. This is in the spirit of discrete mathematics' counting in two ways.

The small blue squares of any individual larger square represent the last term of your sum. Here, I have shown $n=1,2,3,4$. Obviously, each large square has $n^2$ area. Or you can see that it can also be thought of as adding up the "L" shaped additions. Since both of these methods are counting the same thing, they must be the same number.