How to prove that $\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$ for every $k\geqslant1$

Does anyone have any idea how to prove that

$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$

Hint

Notice that

$$\small\frac{1}{n(n+1)(n+2)\cdots(n+k)}=\frac1k\left(\frac{1}{n(n+1)(n+2)\cdots(n+k-1)}-\frac{1}{(n+1)(n+2)\cdots(n+k)}\right)$$ and then telescope.

Here is a way to solve this without telescoping. Define $a_n=\displaystyle\frac{k!}{n(n+1)(n+2)\cdots(n+k)}$. The partial fraction expansion of $a_n$ is $\displaystyle \sum_{m=0}^k\frac{(-1)^m{k\choose m}}{n+m}.$ Let $\displaystyle f(x)=\sum_{m=0}^k\frac{(-1)^m{k\choose m}}{n+m}x^{n+m},$ so that $f(0)=0$, $f(1) = a_n$, and $f'(x)=x^{n-1}(1-x)^k$, by the binomial formula. Thus $a_n=f(1)=\int_0^1f'(x)\,dx=\int_0^1 x^{n-1}(1-x)^k\, dx$. Then

$$ \begin{align*} \sum_{n=1}^\infty \frac{k!}{n(n+1)(n+2)\cdots(n+k)} &= \sum_{n=1}^\infty \int_0^1 x^{n-1}(1-x)^k\,dx.\\ &= \int_0^1 \left(\sum_{n=1}^\infty x^{n-1}(1-x)^{k}\right)\,dx\\ &=\int_0^1(1-x)^{k-1}\,dx\\ &=\frac{1}{k}. \end{align*} $$

Interchanging the summation and integration is valid because of monotone convergence.