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.

I see no one likes induction.
For $n=0$,
$$
\frac{x^0-y^0}{x-y}=\sum_{1 \le i \le 0}x^{0-i}y^{i-1}=0.
$$
Assume for $n=j$ that the identity is true.
Then, for $n=j+1$,
$$
\begin{align}
\sum_{1 \le i \le j+1}x^{j+1-i}y^{i-1}&=\left(\sum_{1 \le i \le j}x^{j+1-i}y^{i-1}\right)+y^j\\
&=\left(x\sum_{1 \le i \le j}x^{j-i}y^{i-1} \right)+y^{j}\\
&=x\frac{x^{j}-y^{j}}{x-y}+y^{j}\\
&=\frac{x(x^j-y^j)+y^j(x-y)}{x-y}\\
&=\frac{x^{j+1}-xy^j+y^jx-y^{j+1}}{x-y}\\
&=\frac{x^{j+1}-y^{j+1}}{x-y}.
\end{align}
$$
Hence, it is as we sought.
