<s>[INST] Write $x^k - y^k$ as the product of two factors. Write $x^k - y^k$ as the product of two factors.
$x^3 - y^3 = (x-y)(x^2+xy+y^2)$
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \cdots + xy^{k-2}}_{k-2} +y^{k-1})$
It's not exact what the middle terms should be, so I'm not sure how to write the expression for the second factor. I think I may need to use the parity of $k$ and some mention of mirroring, though mirroring sounds wrong.
 [/INST] Here is a variation without using induction. We want to use the sigma symbol and start with an example for small $n=4$ to better see what's going on.
We have
\begin{align*}
x^4-y^4&=(x-y)(x^3+x^2y+xy^2+y^3)\\
&=(x-y)(x^3y^0+x^2y^1+x^1y^2+x^0y^3)\tag{1}\\
&=(x-y)\sum_{k=0}^3 x^{3-k}y^k
\end{align*}
Since we can write $x^3=x^3y^0$ and $y^3=x^0y^3$ each term in (1) has a representation $x^{3-k}y^k$, $0\leq k\leq 3$ which can be conveniently written using the sigma symbol $\sum$.

For the general case with integral $n>0$ we have
\begin{align*}
\color{blue}{x^n-y^n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots + xy^{n-2}+y^{n-1})\\
&=(x-y)(x^{n-1}y^0+x^{n-2}y^1+\cdots + x^1y^{n-2}+x^0y^{n-1})\\
&\,\,\color{blue}{=(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}\tag{2}
\end{align*}

We prove (2) algebraically. We obtain for integral $n>0$ by starting with the right-hand side of  (2)
\begin{align*}
\color{blue}{(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}&=x\sum_{k=0}^{n-1} x^{n-1-k}y^k-y\sum_{k=0}^{n-1} x^{n-1-k}y^k\tag{3.1}\\
&=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=0}^{n-1} x^{n-1-k}y^{k+1}\tag{3.2}\\
&=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=1}^{n} x^{n-k}y^{k}\tag{3.3}\\
&\,\,\color{blue}{=x^n-y^n}
\end{align*}
and the claim follows.
Comment:

*

*In (3.1) we multiply out $(x-y)\sum \Box=x\sum \Box - y\sum\Box$.


*In (3.2) we simplify by multiplying with $x$ resp. $y$. Note, we have $x\sum x^{n-1-k}y^k=\sum x^{1+(n-1-k)}y^k=\sum x^{n-k}y^k$.


*In (3.3) we shift the index $k$ in the right-hand side by one. We so have lower bound $k=1$ and upper bound $k=n$. We compensate this shift by substituting $k\to k-1$ within the sum. We now have the convenient situation that we have a telescoping sum with alike terms in the left-hand sum and the right-hand sum for $k=1,\ldots,n-1$, so that everything cancels out besides $x^n-y^n$.
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