Algebraic Identity $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$ Prove the following: $\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$. 

So one could use induction on $n$? Could one also use trichotomy or some type of combinatorial argument?
I have no idea what you mean by "use trichotomy," but here is the combinatorial argument.  $a^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ and $b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... b \}$.  Assume $a &gt; b$.  Then $a^n - b^n$ counts the number of words of length $n$ on the alphabet $\{ 1, 2, ... a \}$ such that at least one letter is greater than $b$.  

Given such a word, suppose the last letter greater than $b$ occurs at position $k+1$.  Then there are $a - b$ choices for this letter, $a^k$ choices for the letters before this letter, and $b^{n-k-1}$ choices for the letters after this letter.  Thus there are $(a - b) a^k b^{n-k-1}$ such words, and summing over all $k$ gives

$$a^n - b^n = (a - b) \sum_{k=0}^{n-1} a^k b^{n-k-1}$$

as desired.
If you know the sum of a geometric sequence, then set $x=a/b$ and conclude
$
x^n-1 = (x-1)(1+x+x^2+\cdots+x^{n-1})
$. Now multiply by $b^n$. (If $b=0$, then the identity is obvious.)