Is this "combinatorial" sum equal to $1$ for every natural $m$? I did some computations and it seems to me that this holds:

$\dfrac {m}{m+1} \cdot \sum_{k=1}^{\infty} \dfrac {1}{m+k \choose m+1}=1$ for every $m \in \mathbb N$.

How to prove this beautiful identity, if it is really true?
 A: It is enough to exploit Euler's Beta function and a geometric series:
$$\begin{eqnarray*}\sum_{k\geq 1}\binom{m+k}{m+1}^{-1} &=& \sum_{k\geq 1}\frac{(m+1)!(k-1)!}{(m+k)!}\\
&=& (m+1)\sum_{k\geq 1}\frac{\Gamma(m+1)\Gamma(k)}{\Gamma(m+k+1)}\\
&=& (m+1) \sum_{k\geq 1} B(k,m+1) \\
&=& (m+1) \sum_{k\geq 1}\int_{0}^{1} x^{k-1}(1-x)^{m}\,dx\\
&=& (m+1) \int_{0}^{1}(1-x)^{m-1}\,dx\\
&=& (m+1) \int_{0}^{1} x^{m-1}\,dx = \color{blue}{\frac{m+1}{m}}.
\end{eqnarray*}$$
A: The proof that this identity holds is easy if you are familiar with Discrete Calculus and negative-exponent falling powers.
In particular, $$\frac{1}{\binom{m+k}{m+1}} = \frac{(m+1)!}{(m+k)^{\underline{m+1}}} = (m+1)! k^{\underline{-(m+1)}} $$
For example, with $m=3,k=7$, 
$$
\frac{1}{\binom{10}{4}} = \frac{4!}{7\cdot 8\cdot 9\cdot 10}
$$
The rules of finite summation work much like those of integration. The "indefinite sum" is
$$
\sum x^{\underline{n}} = \frac{1}{n+1} x^{\underline{n+1}}
$$
(except if $n=-1$ in which case the analogue of $\int x^{-1} dx = \log x$ is $\sum n^{-1} = H_n$, the harmonic sum function).
In particular
$$
\sum k^{\underline{-(m+1)}}  = -\frac{1}{m} k^{\underline{-m}}
$$
and inserting the limits $1$ to $\infty$ the infinity end gives zero and the minus sign goes away leaving 
$$
\sum_{k=1}^\infty k^{\underline{-(m+1)}} = \frac{1}{m} \cdot 1^{\underline{-m}} = \frac{1}{m\cdot m!}$$
So 
$$\frac{1}{\binom{m+k}{m+1}} =(m+1)!\sum \frac{1}{(m+k)^{\underline{m+1}}} = (m+1)! \frac{1}{m\cdot m!} = \frac{(m+1)!}{m!} \frac{1}{m} = \frac{m+1}{m}$$
A: I hope noone minds, if I add an easy approach to this after such a long time. I really enjoyed learning something here from previous answers, but thought a basic approach might be helpful or interesting as well.
$$\sum_{k=1}^n \frac{1}{\binom{m+k}{m+1}}$$$$=\sum_{k=1}^n\frac{(k-1)!(m+1)!}{(m+k)!}$$$$=(m+1)!\sum_{k=1}^n\frac{1}{k\cdot\ldots\cdot(m+k)}$$$$=\frac{(m+1)!}{m}\sum_{k=1}^n\frac{m+k-k}{k\cdot\ldots\cdot(m+k)}$$$$=\frac{(m+1)!}{m}\sum_{k=1}^n\frac{1}{k\cdot\ldots\cdot(m+k-1)}-\frac{1}{(k+1)\cdot\ldots\cdot(m+k)}$$$$=\frac{(m+1)!}{m}(\frac{1}{m!}- \frac{1}{(n+1)\cdot\ldots\cdot(n+k)})$$ and letting n go to infinity brings the result. 
