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

As to "why" $a - b$ divides $a^n - b^n$, here's one attempt to "convince" you. Let $c = a - b$. Then $$a - b \mid a^n - b^n \iff c \mid (b + c)^n - b^n.$$
Expanding $(b + c)^n$ we get
$$(b + c)^n = b^n + \binom{n}{1} b^{n-1} \color{red}{c} + \binom{n}{2} b^{n-2} \color{red}{c^2} + \dots + \binom{n}{n-1} b \color{red}{c^{n-1}} + \color{red}{c^n}.$$
Since all terms are divisible by $c$, except for possibly the first one (which is cancelled by $-b^n$), it follows that $c \mid (b + c)^n - b^n$ for all $b,c$. 
