I was reading about telescoping series and saw this example and was interested in its result. But I was a little confused because some of the steps to the derivation were skipped. In particular, I am interested in knowing why $r$ was removed from $u_{n}$ and how it was decomposed to yield $\frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)}$. Some explanation as to how the final result was determined would also be greatly appreciated. Thank you!
$$\sum_{k=1}^nu_n=\sum_{k=1}^n \frac{1}{k(k+1)...(k+r)} = \frac{1}{r} \sum_{k=1}^n \left( \frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)} \right) = \frac{1}{r} \left( \frac{1}{r!} - \frac{n!}{(n+r)!} \right).$$