<s>[INST] Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
          &= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
          &= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2}  - \cdots- x y^{n-1} - y^n \\
          &= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n  \\
          &\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:


*
  
*Associate law for addition
  
*Existence of an additive identity
  
*Existence of additive inverses
  
*Commutative law for additions
  
*Associative law for multiplication
  
*Existence of a multiplicative identity
  
*Existence of multiplicative inverses
  
*Commutative law for multiplication
  
*Distibutive law

 [/INST] 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...
$$\eqalign{
  & {x^n} - 1 = \left( {x - 1} \right)\sum\limits_{k = 0}^{n - 1} {{x^k}}   \cr 
  & \frac{{{b^n}}}{{{a^n}}} - 1 = \left( {\frac{b}{a} - 1} \right)\sum\limits_{k = 0}^{n - 1} {\frac{{{b^k}}}{{{a^k}}}}   \cr 
  & \frac{{{b^n} - {a^n}}}{{{a^n}}} = \left( {\frac{{b - a}}{a}} \right)\sum\limits_{k = 0}^{n - 1} {\frac{{{b^k}}}{{{a^k}}}}   \cr 
  & {b^n} - {a^n} = \left( {b - a} \right)\sum\limits_{k = 0}^{n - 1} {{b^k}{a^{n - k - 1}}}  \cr} $$
</s>