Here I am again, trying to organize my thought. Thank you Don for encouraging me yesterday. (From this thread Asking for suggestions about square numbers)
I'll try to explain as much as I can, but some parts I really don't know how to explain them...
When you look at the differences between each of the sq number, you find that the differences increase steadily with +2 to each. I call it layers until steady. (Just want some words to identify it).
We can see that numbers of layers agree with number of the ^ (I don't know what is the word for the ^....). See the ^3 layers
Same goes with ^4 and ^5 and above
Now let's take a look at a concrete picture of 3^2 and 4^2
If we want to find their differences, we pull 1 from each ball from 4^2 (to make each ball has 3), we need to pull only 3 balls because we're making them 3^2; and we have one of the 4^2 ball left.
Thus, 4^2 - 3^2 = [3x3](3, which remain in each ball after we pulled 1 out) + 3(pulled from three of the 4^2 balls) + 4 (that ball left from 4^2)
See more picture, please.
I'll continue in here (Continue) About square numbers because I my reputation is not enough to post more than 8 links.
For any sequence $u_n$, define the forward difference $\Delta u_n:=u_{n+1}-u_n$, in analogy with the derivative $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ of a continuous function $f$. Your first finding is $\Delta n^2=2n+1$. We can iterate $\Delta$, e.g. $\Delta^2u_n:=\Delta(\Delta u_n)$ so $\Delta^2n^2=2$. The pattern you've found is $\Delta^kn^k=k!$ for any integer $k\ge0$ (in analogy with $\frac{d^k}{dx^k}x^k=k!$). To prove this, it suffices to note from the binomial theorem that $\Delta n^k$ is a degree-$(k-1)$ polynomial in $n$ with leading coefficient $k$.