$\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)…(k+m+1)}=\frac{1}{(m+1)\cdot(m+1)!}$ Can I proof this identity somehow without using the Riemann integral?
I have tried Stolz–Cesàro Theorem and sums of valuations' progressions.

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$\ds{\lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over k\pars{k + 1}\ldots\pars{k+m+1}}
     ={1 \over m\,m!}:\ {\Large ?}}$

\begin{align}
&\color{#00f}{\large\sum_{k = 1}^{\infty}{1 \over k\pars{k + 1}\ldots\pars{k+m+1}}}=
\sum_{k = 1}^{\infty}{\Gamma\pars{k} \over \Gamma\pars{k + m + 2}}
=
{1 \over \pars{m + 1}!}\sum_{k = 1}^{\infty}
{\Gamma\pars{k}\Gamma\pars{m + 2} \over \Gamma\pars{k + m + 2}}
\\[3mm]&=
{1 \over \pars{m + 1}!}\sum_{k = 1}^{\infty}{\rm B}\pars{k,m + 2}=
{1 \over \pars{m + 1}!}\sum_{k = 1}^{\infty}
\int_{0}^{1}t^{k - 1}\pars{1 - t}^{m + 1}\,\dd t
\\[3mm]&=
{1 \over \pars{m + 1}!}\int_{0}^{1}\pars{1 - t}^{m + 1}
\pars{\sum_{k = 1}^{\infty}t^{k - 1}}\,\dd t
=
{1 \over \pars{m + 1}!}\int_{0}^{1}\pars{1 - t}^{m + 1}\,{1 \over 1 - t}\,\dd t
\\[3mm]&=
{1 \over \pars{m + 1}!}\bracks{-\,{\pars{1 - t}^{m + 1} \over m + 1}}_{0}^{1}
=\color{#00f}{\large{1 \over \pars{m + 1}\pars{m + 1}!}}
\end{align}

$\ds{{\rm B}\pars{x,y} \equiv \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$
is the Beta Function which satisfies
$\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over\Gamma\pars{x + y}}}$. $\Gamma\pars{z}$ is the Gamma Function. Also, $\ds{\Gamma\pars{n + 1} = n!}$
with $\ds{n \in {\mathbb N}}$.
