algebra precalculus - Sum of first n natural numbers proof

I know how to prove this by induction but the text I'm following shows another way to prove it and I guess this way is used again in the future. I'm confused by it.

So the expression for first n numbers is:
$$\frac{n(n+1)}{2}$$

And this second proof starts out like this. It says since:

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

Absolutely no idea where this expression came from, doesn't explain where it came from either.

Then it proceeds to say:

$$\begin{align} 2^2-1^2&=2*1+1 \\ 3^2-2^2&=2*2+1\\ &\dots\\ n^2-(n-1)^2&=2(n-1)+1\\ (n+1)^2-n^2&=2n+1 \end{align}$$
At this point I'm completely lost.
But it continues to say "adding and noting the cancellations on the left, we get"
$$\begin{align} (n+1)^2-1&=2(1+2+...+n)+n \\ n^2+n&=2(1+2+...+n) \\ (n(n+1))/2&=1+2+...+n \end{align}$$

Which proves it but I have no clue what has happened. I am entirely new to these math proofs. Im completely lost. I was great at high school math and calculus but now I haven't got the slightest clue of what's going on. Thanks