Why does $\sum_{k=0}^\infty\frac{r^k(k+n)!}{k!}=\frac{n!}{(1-r)^{n+1}}$? When I put the following series in Mathematica, I get an answer:
$$\sum_{k=0}^\infty\frac{r^k(k+n)!}{k!}=\frac{n!}{(1-r)^{n+1}}$$
Here $0<r<1$ and $n$ is a non-negative integer.
My question is: how does one arrive at this solution (without the use of Mathematica)?  I'd like to learn how to solve this by hand...  The $\frac{(k+n)!}{k!}$ looks like a Pochammer symbol $(k+1)^{(n)}$...

Hint: $\dfrac{(k+n)!}{k!}=n!\cdot\dfrac{(k+n)!}{n!\cdot k!}=n!\cdot\Large{n+k\choose k}$. Now see binomial series.
