Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$

Here is the inductive step, presented more conceptually

$$\rm\frac{x^{n+1}-y^{n+1}}{x-y}\: =\ x^n\: +\ y\ \frac{x^n-y^n}{x-y}$$

So, intuitively, proceeding inductively yields

$$\rm\:x^n + y\: (x^{n-1} + y\: (x^{n-2} +\:\cdots\:))\ =\ x^n + y\: x^{n-1} + y^2\: x^{n-2} + \:\cdots $$

Use this intuition to compose a formal proof by induction.

You have everything right except the last line.

Maybe it is easier to do in this order:

$$(x−y)\left(x^{n−1}+x^{n−2}y+\cdots+xy^{n−2}+y^{n−1}\right)=\\ =x\cdot x^{n-1}-y\cdot x^{n-1} +x\cdot x^{n−2}y- y\cdot x^{n−2}y+x\cdot x^{n−3}y^2-\cdots\\ \cdots -y\cdot x^2y^{n-3} +x\cdot xy^{n-2}-y \cdot y^{n-1}$$

The second term $y\cdot x^{n-1}$ is the same as the third term $x\cdot x^{n−2}y$ except the sign, similarly the 4th and the 5th terms are canceled... So the only terms left are: $x\cdot x^{n-1}$ and $y\cdot y^{n-1}$.

I think it would be easier for you to recall

$$\left(1+x+x^2+\cdots+x^{n-1}\right)(x-1) = x^n-1$$

and put $x=\dfrac{b}{a}$

$$\eqalign{ & \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {\frac{b}{a} - 1} \right) = \frac{{{b^n}}}{{{a^n}}} - 1 \cr & \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {\frac{{b - a}}{a}} \right) = \frac{{{b^n} - {a^n}}}{{{a^n}}} \cr & {a^{n - 1}}\left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {b - a} \right) = {b^n} - {a^n} \cr & \left( {{a^{n - 1}} + b{a^{n - 2}} + {b^2}{a^{n - 3}} + \cdots + {b^{n - 1}}} \right)\left( {b - a} \right) = {b^n} - {a^n} \cr} $$

A little bit "tidier", so that we know what happens in between the dots...

$$\eqalign{ & {x^n} - 1 = \left( {x - 1} \right)\sum\limits_{k = 0}^{n - 1} {{x^k}} \cr & \frac{{{b^n}}}{{{a^n}}} - 1 = \left( {\frac{b}{a} - 1} \right)\sum\limits_{k = 0}^{n - 1} {\frac{{{b^k}}}{{{a^k}}}} \cr & \frac{{{b^n} - {a^n}}}{{{a^n}}} = \left( {\frac{{b - a}}{a}} \right)\sum\limits_{k = 0}^{n - 1} {\frac{{{b^k}}}{{{a^k}}}} \cr & {b^n} - {a^n} = \left( {b - a} \right)\sum\limits_{k = 0}^{n - 1} {{b^k}{a^{n - k - 1}}} \cr} $$