Prove a combinatorial identity: $ \sum_{n_1+\dots+n_m=n} \prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}$ Prove the combinatorial identity
$$
  \sum_{n_1+\ldots+n_m=n} \;\; 
    \prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}, \enspace n_i>0,i=1,\ldots,m
$$
I "discovered" this equality during experiments with Maple, but I have no idea how to prove it. It may have a connection with Catalan numbers but that hasn't helped me.
UPDATE
Now we have brilliant proof of this equality. But may be it have purely combinatorial proof? Or proof with Catalan number's properties?

Using the generating function of the Catalan numbers the left is
$$[z^n] \left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m \; dz.$$
Using Lagrange inversion put $1-4z=w^2$ so that $1/4-z=1/4 \times w^2$
or $z=1/4\times(1-w^2)$ and $dz =  -1/2 \times w\; dw.$ This gives for
the integral
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(-1 +\frac{1-w}{1/2\times(1-w^2)}\right)^m
\times\left(-\frac{1}{2} w\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{w^2-1+2-2w}{1-w^2}\right)^m
\times\left(-\frac{1}{2} w\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\frac{(w-1)^{2m}}{(1-w^2)^{m}}
\times\left(-\frac{1}{2} w\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+m+1}}
(1-w)^{2m}
\times\left(-\frac{1}{2} w\right) \; dw
\\ = - \frac{2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{n+m+1}(1+w)^{n+m+1}}
(1-w)^{2m} \times w\; dw
\\ = - \frac{2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{n-m+1}(1+w)^{n+m+1}}
\times w\; dw
\\ = \frac{(-1)^{n-m} 2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m+1}(1+w)^{n+m+1}}
\times w\; dw.$$
The integral has two pieces, the first is
$$\frac{(-1)^{n-m} 2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m}(2+w-1)^{n+m+1}}
\; dw
\\ = \frac{(-1)^{n-m} 2^{n-m}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m}(1+(w-1)/2)^{n+m+1}}
\; dw.$$
This is
$$(-1)^{n-m} 2^{n-m} 
(-1)^{n-m-1} {n-m-1+n+m\choose n+m}\frac{1}{2^{n-m-1}}
= -2{2n-1\choose n+m}.$$
The second piece is
$$\frac{(-1)^{n-m} 2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m+1}(2+w-1)^{n+m+1}}
\; dw
\\ = \frac{(-1)^{n-m} 2^{n-m}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m+1}(1+(w-1)/2)^{n+m+1}}
\; dw.$$
This is
$$(-1)^{n-m} 2^{n-m} 
(-1)^{n-m} {n-m+n+m\choose n+m}\frac{1}{2^{n-m}}
= {2n\choose n+m}.$$
Collecting the two pieces we obtain
$${2n\choose n+m}
-2{2n-1\choose n+m}
= {2n\choose n+m} - 2\frac{n-m}{2n}{2n\choose n+m}
\\= \frac{2n-2n+2m}{2n}{2n\choose n-m}
= \frac{m}{n}{2n\choose n-m}.$$
