Picture proof for expansion of $x^n−y^n$

Here's a semi-visual, semi-algebraic approach.

A useful alternative perspective on (univariate) polynomials is to think of them as being represented by their sequence of coefficients – filling in zeroes for any missing monomials. With this perspective, multiplying by $z$ simply shifts the sequence over by one. Arranging the coefficients in a table makes the result obvious. Here it is for $n=3$ using the $z$ form.

	0	1	2	3	Corresponding polynomial
(1)	$1$	$1$	$1$	$0$	$z^2 + z + 1$
(2) = $z$ times (1)	$0$	$1$	$1$	$1$	$z(z^2 + z + 1)$
(3) = $-1$ times (1)	$-1$	$-1$	$-1$	$0$	$-(z^2 + z + 1)$
(4) = (2) + (3)	$-1$	$0$	$0$	$1$	$z^3 - 1$

This is, of course, the typical algebraic proof just written with a different notation for polynomials, but I think it makes it pretty obvious how to arrive at the solution and why it is correct and will generalize. I do find viewing polynomials as their sequences of coefficients does often make what is happening clearer. Theoretically, a benefit of defining operations in terms of this sequence perspective, is that you are always operating on a normal form representation of the polynomials, so you never end up with the same polynomial presented in different ways.

$$\begin{array}{c} \times & \mid & x^{n−1} & +x^{n−2}y & +\ldots & +xy^{n−2} & +y^{n−1} \\ \hline x & \mid & x^n & {\color{blue} +x^{n−1}y} & +\ldots & +x^2 y^{n−2} & {\color{green} +x y^{n−1}} \\ -y & \mid & {\color{blue} -x^{n−1}y} & -x^{n−2}y^2 & -\ldots & {\color{green} -xy^{n−1}} & -y^n \end{array}$$