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}$

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 $${2n\choose n+m} -2{2n-1\choose n+m} = {2n\choose n+m} - 2\frac{n-m}{2n}{2n\choose n+m} \\= \frac{2n-2n+2m}{2n}{2n\choose n-m} = \frac{m}{n}{2n\choose n-m}.$$

Since $$ \sum_{n\geq 1}\frac{1}{n}\binom{2n}{n-1}x^n = -1+\frac{1-\sqrt{1-4x}}{2x}=\frac{1-\sqrt{1-4x}}{1+\sqrt{1+4x}}\tag{1}$$ the LHS is just the coefficient of $x^n$ in the Taylor series of $$ f(x) = \left(\frac{1-\sqrt{1-4x}}{1+\sqrt{1+4x}}\right)^m \tag{2}$$ that can be computed by using Chebyshev polynomials or the Lagrange inversion theorem, since the inverse function of $\frac{1-\sqrt{1-4x}}{1+\sqrt{1+4x}}$ is just $\frac{y(1-y^2)}{(1+y^2)^2}$.