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.

Yes there is. First let's look at the series:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} \, . $$
This is a geometric series with $n$-terms, whose first term is $x^{n-1}$ and whose common ratio is $y/x$. Applying the standard formula we obtain:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} = x^{n-1}\left( \frac{1-(y/x)^n}{1-(y/x)} \right) . $$
Simplifying the numerator and denominator and then cancelling common factors gives:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} = \frac{x^n-y^n}{x-y} \, . $$
