Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+v)}=\frac{1}{vv!}$$
I am struggling to find a solution for this but no luck yet. How can I analyze it to get to second part?
 A: \begin{eqnarray}
\frac{1}{n(n+1)...(n+v)}&=&\frac{1}{v}\frac{(n+v)-n}{n(n+1)...(n+v)}\\
&=&\frac{1}{v}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
\end{eqnarray}
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)...(n+v)}
=\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]$$
$$\frac{1}{v}\sum_{n=1}^{k}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
=\frac{1}{v}[\frac{1}{v!}-\frac{1}{(k+1)...(k+v)}]$$
Let $k\rightarrow\infty$, then 
$\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
=\frac{1}{v}\frac{1}{v!}$
A: Let $n^{\overline{r}}=\underbrace{n(n+1)(n+2)\cdots (n+r-1)}_{r\text{ terms}}$.
Hence
$$\begin{align}\require{cancel}
&\sum_{n=1}^{\infty}\frac1{n(n+1)(n+2)\cdots (n+v)}\\
&=\sum_{n=1}^{\infty}\frac1{n(n+1)^{\overline{v}}}=\sum_{n=1}^{\infty}\frac1{n^\overline{v}(n+v)}\\
&=\frac1v\sum_{n=1}^{\infty}\frac1{(n)^\overline{v}}-\frac1{(n+1)^{\overline{v}}}\\
&=\frac1v\left[\left(\frac1{1^\overline{v}}-\cancel{\frac1{2^\overline{v}}}\right)+\left(\cancel{\frac1{2^\overline{v}}}-\bcancel{\frac1{3^\overline{v}}}\right)+\left(\bcancel{\frac1{3^\overline{v}}}-\cancel{\frac1{4^\overline{v}}}\right)+\cdots\right]\\
&=\frac1v \left[\frac1{1^\overline{v}}\right]\\
&=\frac1{vv!}\qquad\blacksquare\end{align}$$
