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

A simple way to "prove" the formula you give is to note that you can pair 
$1$ with $n-1$
$2$ with $n-2$
$3$ with $n-3$
... and so on. Then you can see that the sum will be $(n+1)/2$ (the $+1$ coming from the fact that we basically pair $n$ with zero) of these pairs, each of which is the number $n$. 
