Identity proof $(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$ In a proof from a textbook they use the following identity (without proof):

$(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$

Is there an easy way to prove the above? I suppose maybe an induction proof will be appropriate, but I would really like to find a more intuitive proof.

My pre-calc instructor showed me this way (it's essentially the same as some of the answers above, but the details are a little bit different):
Define $S_n:= x^{n-1} + yx^{n-2} + y^2x^{n-3}+...+ \ y^{n-3}x^2 + y^{n-2}x + y^{n-1}$
Note that $S_n$ as you've defined it is the same as I've defined it, i.e. $S_n=\sum_{k=1}^{n} x^{n-k}y^{k-1}$. Then consider the following difference:
$(x-y)S_n = \ xS_n - y\ S_n$ = $(x^{n} + yx^{n-1} + y^2x^{n-2}+...+ \ y^{n-3}x^3 + y^{n-2}x^2 + xy^{n-1})-(x^{n-1}y + y^2x^{n-2} + y^3x^{n-3}+...+ \ y^{n-2}x^2 + y^{n-1}x + y^{n})$
This term on the right telescopes - everything in that large term on the right hand side cancels except for $x^n-y^n$. So we can write 
$(x-y)S_n = x^n-y^n$ 
or
$S_n=\frac{x^n-y^n}{x-y}$, and you're done!
