Can someone show me why this factorization is true?

$$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$

Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this

It’s easier to verify that it multiplies back together correctly:

$$x\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)=x^n+\color{red}{x^{n-1}y+\ldots+x^2y^{n-2}+xy^{n-1}}\;,\tag{1}$$

and

$$y\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)=\color{red}{x^{n-1}y+x^{n-2}y^2+\ldots+xy^{n-1}}+y^n\;.\tag{2}$$

The two red blocks are identical, so when you subtract $(2)$ from $(1)$, all that remains is $x^n-y^n$.

Note that the identity has a very simple form in summation notation:

$$x^n-y^n=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k\;.$$

In each term the exponents of $x$ and $y$ sum to $n-1$, and there is a term for every possible pair of non-negative exponents with sum $n-1$. This observation may make it easier to remember.

That said, it is possible to do the long division. Here’s how it starts, enough to show the pattern:

$$\begin{array}{c|ll} &x^{n-1}&+&x^{n-2}y&+&x^{n-3}y^2&+&\dots&+&y^{n-1}\\ \hline x-y&x^n&&&&&&\dots&&&-&y^n\\ &x^n&-&x^{n-1}y\\ \hline &&&x^{n-1}y&&&&&&&-&y^n\\ &&&x^{n-1}y&-&x^{n-2}y^2\\ \hline &&&&&x^{n-2}y^2&&&&&-&y^n\\ &&&&&x^{n-2}y^2&-&x^{n-3}y^3\\ \hline &&&&&&&x^{n-3}y^3\\ &&&&&&\vdots\\ \hline &&&&&&&&&xy^{n-1}&-&y^n\\ &&&&&&&&&xy^{n-1}&-&y^n\\ \hline \end{array}$$

I don't know if you would allow this (If you don't allow it, then I can include a proof of the geometric sum formula).

$(x/y)^n-1=\sum_{k=0}^{n-1}(x/y)^k$ (Geometric sum)

Now multiply by $y^n$, to get:

$x^n-y^n=\sum_{k=0}^{n-1}x^ky^{n-k}$

Since you said above that you already believe that the result is true, that makes me think you really want to know how to come up with the result (if it had not been presented to you).

My answer would be: play with several examples!

That is, sharpen your pencil and get busy factoring out $x^n-y^n$ for $n=1,2,3,\dots$ using long division or any other tools at your disposal. Then look at the pattern formed by the coefficients as well as the powers of $x$ and $y$:

\begin{array}{c|c} n & x^n-y^n \\\hline 1 & x-y \\ 2 & (x-y) (x+y) \\ 3 & (x-y) \left(x^2+y x+y^2\right) \\ 4 & (x-y) \left(x^3+y x^2+y^2 x+y^3\right) \\ 5 & (x-y) \left(x^4+y x^3+y^2 x^2+y^3 x+y^4\right) \\ 6 & (x-y) \left(x^5+y x^4+y^2 x^3+y^3 x^2+y^4 x+y^5\right) \\ 7 & (x-y) \left(x^6+y x^5+y^2 x^4+y^3 x^3+y^4 x^2+y^5 x+y^6\right) \\ 8 & (x-y) \left(x^7+y x^6+y^2 x^5+y^3 x^4+y^4 x^3+y^5 x^2+y^6 x+y^7\right) \\ 9 & (x-y) \left(x^8+y x^7+y^2 x^6+y^3 x^5+y^4 x^4+y^5 x^3+y^6 x^2+y^7 x+y^8\right) \\ 10 & (x-y) \left(x^9+y x^8+y^2 x^7+y^3 x^6+y^4 x^5+y^5 x^4+y^6 x^3+y^7 x^2+y^8 x+y^9\right) \\ \end{array}

Hopefully this would lead you to conjecture that the formula you gave is indeed true for all $n$. Then to prove this, you would proceed by induction.

Hope that helps.