Not sure if it's original, but interesting to think!

$\large \displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)...(n+k-1)(n+k)}$

$\large \displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)...(n+k-1)(n+k)} \color{#3D99F6} \cdot \frac{(n+k) - n}{k}$

$\large \displaystyle S = \frac1k \sum_{n=1}^{\infty} \frac{1}{n(n+1)...(n+k-1)} - \frac{1}{(n+1)(n+2)...(n+k)}$

This will telescope, and only the first term will remain:

$\large \displaystyle S = \frac1k \left [ \frac{1}{n(n+1)(n+2)...(n+k-1)} \right]_{n=1}$

$\color{#3D99F6} \boxed{ \large \displaystyle S = \frac{1}{k\cdot k!} }$

$t_n=\dfrac{1}{n(n+1)(n+2)....(n+k)}=\dfrac{(n-1)!}{(n+k)!}=\dfrac{1}{k!} \dfrac{(n-1)! k!}{(n+k)!} \\ =\dfrac{\Gamma(n) \Gamma(k+1)}{\Gamma(n+k+1)} =\dfrac{1}{k!} B(n,k+1)$

We need to evaluate

$S=\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{k!} B(n,k+1) \\ =\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{k!} \int_{0}^{1} x^{n-1} (1-x)^k dx$

$=\dfrac{1}{k!} \displaystyle \int_{0}^{1} (1-x)^k (\displaystyle \sum_{n=1}^{\infty} x^{n-1}) dx \\ =\dfrac{1}{k!} \displaystyle \int_{0}^{1} (1-x)^k (\dfrac{1}{1-x}) dx$ .......................As $|x|<1$

$=\dfrac{1}{k!} \displaystyle \int_{0}^{1} (1-x)^{k-1} dx \\ =\dfrac{1}{k.k!}$