Spivak's Calculus - Chapter 1 Question 1.5 - Proof by Induction In Spivak's Calculus Fourth Edition, Chapter 1 Question 1.5 is as follows:

Prove $x^n - y^n = (x - y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1})$ using only the following properties:



It's easy enough for me to use P9 to expand the right-hand side:
$$
\begin{array} { c l }
(x - y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1}) & \text{Given} \\
x^{n-1}(x - y)+x^{n-2}y(x - y)+ \cdots + xy^{n-2}(x - y)+y^{n-1}(x - y) & \text{P9} \\
x^n - yx^{n-1} + x^{n-1}y - x^{n-2}y^2 + \cdots + xy^{n-1} - xy^{n-2}y^2 + xy^{n-1} - y^n & \text{P9}
\end{array}
$$
Finally, all the terms except the first and last cancel out by P3.
I have a feeling this last step should be proved by induction. If so--how would one write that out?
( Also, I've noticed that a lot of people write the application of P9 to polynomials differently than I do when they're working on Spivak's Calculus. Why do they do that? What would it look like here? )

The basic steps of proof by induction are:


*

*Identify the Induction Hypothesis
In this case, it is $P(n) : x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})$
Note that it is often a good idea to write the Induction Hypothesis using iterative operators, like $\sum$ and $\prod$; doing so usually makes the analysis easier.
$$P(n) : x^n - y^n = (x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i$$
The $P(n) : $ here reads "The proposition over $n$ that the following is true".

*Identify the smallest value of $n$ for which the induction hypothesis is true.
In this case, $n = 1$

*Assume the Induction Hypothesis is true.

*Take the so-called "induction step". That is, increment $n$ from the Induction Hypothesis once--and show that the resulting proposition is still true.
What does this accomplish?
The above steps are simply a formal way of saying, "If Step 3 is true, Step 4 shows that we can increment $n$ as much as we like--and Step 3 will continue to be true. Because we can start at some minimum value of $n$ and increment it until we've visited every possible value of $n$--and because we've already shown that Step 2 ( the minimum value of $n$ ) is true--we know that Step 3 is true for all $n$
Applying those steps to the original question
Step 1:
$$P(n) : x^n - y^n = (x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i$$
Step 2:
$$P(1) : x^1 - y^1 = (x-y)\sum_{i=0}^{1-1}x^{1-1-i}y^i$$
$$P(1) : x - y = (x-y)\sum_{i=0}^{0}x^{-i}y^i$$
$$P(1) : x - y = (x-y)(x^{-0}y^0)$$
Note that $x^{-0}$ is defined and $x^{-0} = x^0 = 1$. So, by substitution:
$$
P(1) : x - y = (x-y)(1 \cdot 1) \\
P(1) : x - y = (x-y)(1) \\
P(1) : x - y = (x-y) \\
P(1) : true \\
$$
Step 3:
$$P(n) : x^n - y^n = (x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i \text{ is true by assumption.}$$
Step 4:
$$P(n+1) : x^{n+1} - y^{n+1} = (x-y)\sum_{i=0}^{n+1-1}x^{n+1-1-i}y^i \text{ By Induction Hypothesis}$$
$$P(n+1) : x^{n+1} - y^{n+1} = (x-y)\sum_{i=0}^{n}x^{n-i}y^i$$
$$P(n+1) : x^{n+1} - y^{n+1} = (x-y)(x^n+x^{n-1}y+x^{n-2}y^2+\cdots+x^2y^{n-2}+xy^{n-1}+y^n)$$
$$P(n+1) : x^{n+1} - y^{n+1} = x^n(x-y)+x^{n-1}y(x-y)+x^{n-2}y^2(x-y)+\cdots+x^2y^{n-2}(x-y)+xy^{n-1}(x-y)+y^n(x-y)$$
$$P(n+1) : x^{n+1} - y^{n+1} = x^{n+1} - y^{n+1}$$
$$QED$$
Each operation in the above proof can be justified using P1 through P12.
