Derivation of factorization of $a^n-b^n$

We answer the divisibility question, and not the exact form of the quotient, which has already been dealt with. Let $b$ be fixed, and consider the polynomial $P(x)=x^n-b^n$. Then there is a polynomial $Q(x)$, and a constant $r$, such that $$P(x)=(x-b)Q(x)+r.$$ Put $x=b$. Then since $P(b)=0$, and $(b-b)Q(b)=0$, we conclude that $r=0$. It follows that $x^n-b^n=(x-b)Q(x)$ for all $x$. Now put $x=a$.

I'm not sure why this never occurred to me before, but looking at your general formula written out in full, I thought, "Where have we seen something like this before?" We can write

$$ a^n \ - \ b^n \ = \ a^n \ \left[ \ 1 \ - \ \left( \frac{b}{a} \right)^n \ \right] $$

$$ = \ \left[ \ 1 \ - \ \left( \frac{b}{a} \right) \ \right] \ \sum_{k=0}^{n-1} \ a^n \cdot \left( \frac{b}{a} \right)^k $$

$$ = \ \left[ \ 1 \ - \ \left( \frac{b}{a} \right) \ \right] \ \cdot \ a \ \cdot \ a^{n-1} \ \cdot \ \left( \ 1 \ + \ \frac{b}{a} \ + \ \frac{b^2}{a^2} \ + \ \ldots \ + \ \frac{b^{n-1}}{a^{n-1}} \ \right) $$

$$ = \ (a - b) \ ( \ a^{n-1} \ + \ a^{n-2}b \ + \ a^{n-3}b^2 \ + \ \ldots \ + \ b^{n-1} \ ) \ \ . $$

This also suggests a variant proof analogous to the usual one for the "finite geometric series formula":

$$ s \ = \ 1 \ + \ \frac{b}{a} \ + \ \frac{b^2}{a^2} \ + \ \ldots \ + \ \frac{b^{n-1}}{a^{n-1}} $$

$$ \underline{- \ \ \left(\frac{b}{a} \right) \ s \ = \ \frac{b}{a} \ + \ \frac{b^2}{a^2} \ + \ \ldots \ + \ \frac{b^{n-1}}{a^{n-1}}\ + \ \frac{b^n}{a^n} }$$

$$ \left[ \ 1 \ - \ \left(\frac{b}{a} \right) \ \right] \ s \ = \ 1 \ - \ \frac{b^n}{a^n} \ , $$

with multiplication of both sides by $ \ a^n \ $ carried through in the same manner as in the derivation above:

$$ a \ \left[ \ 1 \ - \ \left(\frac{b}{a} \right) \ \right] \ \cdot \ a^{n-1} \ \cdot \ s \ = \ a^n \ - \ b^n \ \ . $$

[This is related to the expression Git Gud uses, but this is a proof without the use of induction.]

See what happens when you multiply the second factor with the first one. When you multiply it by $a$ make not of all terms except $a^n$. Then multiply the second factor by $(-b)$. You will not that every element except the last one($b^n$) cancels out. This is why the result is true.

For the divisibility problem you also need the condition that $a, b \in \Bbb Z$. By definition $r$ divides $s$ iff $s = kr$ for some integer $k$. As long as $a$ and $b$ are integers the second factor too is an integer and we have that $(a - b)$ times another integer is equal to $a^n - b^n$ and hence divisibility follows.

Hope I helped.