Derivation of factorization of $a^n-b^n$ How does one prove that:
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$$
Better yet, why is $a^n-b^n$ divisible by $a-b$? I would very much appreciate some help on this. Thanks

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.]
