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? 

Can we build a combinatorial argument along these lines?
Say if we have say $n$ students to be allotted in $a$ rooms with $b$ of the $a$ room being non air-conditioned. (Assume the students are distinguishable so we can order the student as $1,2,3,...,n$)
The total number of ways is $a^n$.
Suppose all students are allotted to the $b$ rooms, the number of ways is $b^n$.
If the first $n-1$ students are allotted to the $b$ rooms, and the final dude in some other room, the number of possible ways is $b^{n-1} \times (a-b)$.
Now if the first $n-2$ students are allotted to the $b$ rooms, and the remaining two students are now left. If the $(n-1)^{th}$ student chooses from the $b$ rooms then we are back to the earlier case. So the $(n-1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the $n^{th}$ student can choose from any of the $a$ rooms. The number of possible ways is $b^{n-2} \times (a-b) \times a$.
In general, if the first $n-k$ students are allotted to the $b$ rooms, and the remaining $k$ students are now left. If the $(n-k+1)^{th}$ student chooses from the $b$ rooms then we are back to the previous case. So the $(n-k+1)^{th}$ student needs to choose from the remaining $(a-b)$ rooms. Now the students from $(n-k+2)$ to $n$ can choose from any of the $a$ rooms. The number of possible ways is $b^{n-k} \times (a-b) \times a^{k-1}$.
So, the total number of ways is $b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.
Both the counting must add up and hence we get $a^n = b^n + \displaystyle \sum_{k=1}^n b^{n-k} \times (a-b) \times a^{k-1}$.
You could use geometric series to conclude the result as well.
The right hand side is $(a-b) b^{n-1} \displaystyle \sum_{k=0}^{n-1} (\frac{a}{b})^k = (a-b) b^{n-1} \frac{((\frac{a}{b})^n - 1)}{(\frac{a}{b})-1} = (a-b) \frac{a^n - b^n}{a-b} = a^n - b^n$
