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