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.

The sum you bracketed can be expressed as $$x^{k-2}y+\cdots +xy^{k-2} =\sum_{i=1}^{k-2}x^{k-2+1-i}y^i =\sum_{i=0}^{k-1}x^iy^{k-j-1} -x^{k-1}-y^{k-1}$$ For a formal proof: Notice that $$a^2-b^2 = a^2+ab-ab-b^2 = a(a+b)-b(a+b) = (a-b)(a+b)$$ so that our result holds for $n=2$.Now suppose that $$a^k -b^k = (a-b)(a^{k-1}+ba^{k-2}+b^2a^{k-3}+b^3a^{k-4} +\cdots +b^{k-3}a^2 +b^{k-2}a +b^{k-1})$$ for some $k\in \mathbf{N}$ with $k\geqslant 2.$ Then \begin{align*}
a^{k+1}-b^{k+1} & = aa^k-bb^k \\ &= aa^k-ba^k+ba^k-bb^k \\ &= a^k(a-b)+b(a^k-b^k) \\ &= a^k(a-b)+b\left((a-b)(a^{k-1}+ba^{k-2}+\cdots +b^{k-2}a +b^{k-1})\right) \hspace{8mm}\text{Inductive Hypothesis}\\ &= (a-b)(a^k +b(a^{k-1}+ba^{k-2}+\cdots +b^{k-2}a +b^{k-1}))\\ &= (a-b)(a^{k}+ba^{k-1}+b^2a^{k-2}+b^3a^{k-3} +\cdots +b^{k-2}a^2 +b^{k-1}a +b^{k}).
\end{align*}
Therefore, by the Principles of Mathematical Induction, we have that $$a^n -b^n = (a-b)(a^{n-1}+ba^{n-2}+b^2a^{n-3}+b^3a^{n-4} +\cdots +b^{n-3}a^2 +b^{n-2}a +b^{n-1})$$ for any $a$, $b\in \mathbf{R}$ and $n\in \mathbf{N}.\blacksquare$
