Find $\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$ Original question is $$\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$$
I got it down to $$\sum_{n=1}^{\infty} \frac{(n-1)!}{(k+n)!}$$
Here I am confused.  Possible fraction decomposition but its ugly!  Maybe this approach is not good?  Ideas?
Answer is $$\frac{1}{k \cdot k!}$$
I want to know how to proceed with my work though

\begin{align*}
\sum_{n=1}^\infty \frac{1}{n(n+1)...(n+k)} &= \frac{1}{k} \sum_{n=1}^\infty \frac{k}{n(n+1)...(n+k)} \\
 &= \frac{1}{k} \sum_{n=1}^\infty \left[ \frac{1}{n(n+1)...(n+k-1)} - \frac{1}{(n+1)...(n+k)} \right], \\
\end{align*}
and this series telescopes so that every minus cancels with a plus, and we are left with only the first plus term, when $n = 1$:
$$\frac{1}{k} \frac{1}{1(1+1)...(1+k-1)} = \frac{1}{k * k!}$$
