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


Since $x^{n-1} + x^{n-2} y + \dots + x y^{n-2} + y^{n-1}$ is a geometric series with $n$ terms and a common factor of $y/x$, it equals
$$
\frac{x^{n-1}\left(1-\left(\frac{y}{x}\right)^n\right)}{1-\frac{y}{x}}=\frac{x^{n}\left(1-\left(\frac{y}{x}\right)^n\right)}{x-y}=\frac{x^n-y^n}{x-y} \, ;
$$
multiplying through by $(x-y)$ gives the desired result.
