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? 
 A: 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
$x^n-a^n\equiv (x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots +a^{n-2}x+a^{n-1})$
A: EDIT
Proof by induction
$n=1$ is valid.
Supose valid by n, then
$$a^{n+1}-b^{n+1}=a(a^{n})+b(b^{n})$$, using the hipotesis :
$$a(a^{n})+b(b^{n})=a\left[b^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + b\left[a^{n}-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$
$$\left[ab^{n}+(b-a)a\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right] + \left[a^{n}b-(b-a)b\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}\right]=$$
$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}\right] + \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}\right]=$$
$$\left[ab^{n}+(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+(b-a)b^{n}-(b-a)b^{n}\right] +$$
$$ \left[a^{n}b-(b-a)\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}-(b-a)a^{n}+(b-a)a^{n}\right]=$$
$$\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k+1}b^{n-1-k}+b^{n}]+b^{n+1}\right] +\left[(b-a)[\displaystyle\sum\limits_{k=0}^{n-1} a^{k}b^{n-k}+a^{n}]-a^{n+1}\right] +$$
$$\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}+b^{n+1}\right] +\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}-a^{n+1}\right] =$$
$$-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] =$$
Thus:
$$a^{n+1}-b^{n+1}=-a^{n+1}+b^{n+1}+2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right] $$, then 
$$2(a^{n+1}-b^{n+1})=2\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$
thus:
$$a^{n+1}-b^{n+1}=\left[(b-a)\displaystyle\sum\limits_{k=0}^{n} a^{k}b^{n-k}\right]$$
So $n+1$ is valid.
Complete the proof
A: 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 $(a-b) b^{n-1} \displaystyle \sum_{k=0}^{n-1} (\frac{a}{b})^k = (a-b) b^{n-1} \frac{((\frac{a}{b})^n - 1)}{(\frac{a}{b})-1} = (a-b) \frac{a^n - b^n}{a-b} = a^n - b^n$
A: Yes, you could use induction on n.  I don't see an easy trichotomy or combinatorial argument.
A: 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 $a > b$.  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.
A: Someone should mention the "polynomial multiplication" or "telescoping" proof, which may be viewed as a variant of the "geometric series" method.
$$\begin{align*} (a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k} &= \sum_{k=0}^{n-1} a^{k+1} b^{n-1-k} - \sum_{k=0}^{n-1} a^k b^{n-k} \\ &= \sum_{k=1}^n a^k b^{n-k} - \sum_{k=0}^{n-1} a^k b^{n-k} = a^n - b^n. \end{align*}$$
A: 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.)
