Non-Inductive Proof

Proof: Suppose there exists $z$ such that $$x^n-y^n = z(x-y).$$ This would imply that $z=\dfrac{x^n-y^n}{x-y}$. Now, here comes the trick: $$\frac{x^n-y^n}{x-y} = \frac{x^{n-1}(x-y)+yx^{n-1}-y^n}{x-y}=x^{n-1}+y\left(\frac{x^{n-1}-y^{n-1}}{x-y}\right).$$ Continuing in the same way, it follows that $$x^{n-1}+y\left(\frac{x^{n-1}-y^{n-1}}{x-y}\right)=x^{n-1}+yx^{n-2}+y\left(\frac{x^{n-2}-y^{n-2}}{x-y}\right)=\cdots$$ to which a pattern can be noticed, namely, $$z=\sum_{k=1}^nx^{n-k}y^{k-1}\tag*{$\bigcirc$}$$

I decided to not do an inductive proof in order to explore a different way of tackling this problem :)

Non-Inductive Proof (or so I thought).

Proof: Suppose there exists $z$ such that $$x^n-y^n = z(x-y).$$ This would imply that $z=\dfrac{x^n-y^n}{x-y}$. Now, here comes the trick: $$\frac{x^n-y^n}{x-y} = \frac{x^{n-1}(x-y)+yx^{n-1}-y^n}{x-y}=x^{n-1}+y\left(\frac{x^{n-1}-y^{n-1}}{x-y}\right).$$ Continuing in the same way, it follows that $$x^{n-1}+y\left(\frac{x^{n-1}-y^{n-1}}{x-y}\right)=x^{n-1}+yx^{n-2}+y\left(\frac{x^{n-2}-y^{n-2}}{x-y}\right)=\cdots$$ to which a pattern can be noticed, namely, $$z=\sum_{k=1}^nx^{n-k}y^{k-1}\tag*{$\bigcirc$}$$

I decided to not do an inductive proof in order to explore a different way of tackling this problem :)

Edit: Turns out, it is an inductive proof, but it just skips the base case and is differently worded than usual. Credit to @J.G. who pointed that out :)