Combinatorial proof of $a^n - b^n = (a - b)(a^{n – 1} + a^{n – 2}b + \dots + ab^{n – 2} + b^{n – 1})$ Is it possible to come up with a combinatorial argument which proves the following identity?

$$a^n - b^n = (a - b)(a^{n – 1} + a^{n – 2}b + \dots + ab^{n – 2} + b^{n – 1})$$

My idea was this: 

Consider the number of ways $N$ to distribute $n$ numbered balls into $a$ numbered boxes such that there must be at least $1$ ball among the first $a - b$ boxes. The number of ways to do this is take the total number of ways without restriction, and subtract off the number of ways that all the balls are distributed only among the last $b$ boxes. That is,

$$N = a^n - b^n$$

Alternatively, we can first pick at least $1$ ball(s) and distribute them among the first $a - b$ boxes, and then distribute the remaining balls among the last $b$ boxes. That is,

$$N = \binom{n}{1}(a-b)b^{n-1} + \binom{n}{2}(a-b)^2b^{n-2} + \dots + \binom{n}{n}(a-b)^nb^0$$

However, this does not resemble the required $RHS$ of the identity.
Let ball $k$ be the highest numbered ball that gets placed in one of the first $a-b$ boxes.  Then all balls numbered $k+1$ through $n$ get placed in the last $b$ boxes, whereas balls numbered $1$ through $k-1$ may be placed in any of the $a$ boxes.
Using the binomial formula:

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

And 
$$\sum_{k=0}^{n-1}{n\choose k}(a-b)^{n-k-1}b^k = \sum_{k=0}^{n-1} \sum_{j=0}^{n-k-1}{n\choose k}{n-k-1\choose j}(-1)^ja^{n-k-1-j}b^{j+k} = \sum_{i=0}^{n-1}(\sum_{j+k=i} {n\choose k}{n-k-1\choose j}(-1)^j)a^{n-1-i}b^i$$

We need to prove $\sum_{j+k=i} {n\choose k}{n-k-1\choose j}(-1)^j = 1$ i.e. $ \sum_{j=0}^i{i\choose j}\frac{(-1)^j}{n-(i-j)} = \dfrac{1}{{n \choose i}(n-i)}$

This can be proven by partial fraction decomposition combined with the fact $\sum_{j=0}^n(-1)^{n-j}{n\choose j} = 0$.

The partial decomposition says 
\begin{align}
\dfrac{1}{{n\choose i}} &amp;= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-j}{n-j} \\
&amp;=  -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-n}{n-j} - \sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{n-j}{n-j} \\
&amp;= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-n}{n-j} \\
\end{align}
which gives the conclusion
\begin{align}
\dfrac{1}{{n\choose i}(n-i)} = \sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{1}{n-j} = \sum_{j=0}^{i}(-1)^{j}{i \choose j}\frac{1}{n-(i-j)}
\end{align}

Clearly, if we do things in the inverse direction, we can prove the the partial fraction decomposition with $(a^n -b^n)= (a-b)(a^{n-1} + \cdots + b^{n-1})$.