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? 

You can apply Ruffini's rule. Here is a copy from my Algebra text book (Compêndio de Álgebra, VI, by Sebastião e Silva and Silva Paulo) where the following formula is obtained:
$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1}).$

Translation: The Ruffini's rule can be used to find the quotient of $x^n-a^n$ by $x-a$:
(Figure)
Thus, if $n$ is a natural number, we have
$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1})$
