Difference of Two Powers/Proof 4

Proof
The proof will proceed by the Principle of Complete Finite Induction on $\Z_{>0}$.

Let $S$ be the set defined as:
 * $\ds S := \set {n \in \Z_{>0}: a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j}$

That is, $S$ is to be the set of all $n$ such that:
 * $\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$

Basis for the Induction
We have that:


 * $\ds a^1 - b^1 = \paren {a - b} \sum_{j \mathop = 0}^0 a^{1 - 0 - 1} b^j$

So $1 \in S$.

This is the basis for the induction.

Induction Hypothesis
It is to be shown that if $r \in S$ for all $r$ such that $1 \le r \le k$, then it follows that $k + 1 \in S$.

This is the induction hypothesis:
 * $\forall r \in \Z_{>0}: 1 \le r \le k: \ds a^r - b^r = \paren {a - b} \sum_{j \mathop = 0}^{r - 1} a^{r - j - 1} b^j$

It is to be demonstrated that it follows that:
 * $\ds a^{k + 1} - b^{k + 1} = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j$

Induction Step
This is the induction step:

So $\forall r \in S: 0 \le r \le k: r \in S \implies k + 1 \in S$ and the result follows by the Principle of Complete Finite Induction:


 * $\forall n \in \Z_{>0}: \ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$