How to prove $[x^n]C^m(n)=\binom{2n+m-1}{n}-\binom{2n+m-1}{n-1}$ 
Recently I founded that
  $$
[x^n]C^m(x)=\binom{2n+m-1}{n}-\binom{2n+m-1}{n-1}
$$
  where $C^m(x)$ means the $m$-th power of the generating function of Catalan numbers, that is
  $$
C(x)=\frac{2}{\sqrt{1-4 x}+1}
$$

If you prove from right to left, then it will be obvious:
$$
\sum_{n=0}^\infty \left (\binom{2n+m-1}{n}-\binom{2n+m-1}{n-1}\right ) x^n = \, _2F_1\left(\frac{m}{2},\frac{m+1}{2};m;4 x\right)-x \, _2F_1\left(\frac{m+2}{2},\frac{m+3}{2};m+2;4 x\right)=2^m \left(\sqrt{1-4 x}+1\right)^{-m}
$$

However I want to assume that I don't know the right form, that means I want to calculate the left one.

And I found that one possible method is to use Faa di Bruno formula and Bell Polynomials.
My question: Is there any other approaches? The approach above is too troublesome, it indeed needs a lot of calculation. I am wondering if there's more elementary(maybe tricky?) method.
 A: The generating function $C(z)$ of the Catalan numbers is
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}$$
and satisfies the functional equation
$$C(z) = 1+ z C(z)^2 \quad\text{or}\quad
z = \frac{C(z)-1}{C(z)^2}.$$
Introducing
$$C_{n,m} = [z^n] C(z)^m$$
we have
$$ [z^{n-1}] (C(z)^m)' = n C_{n,m} =
[z^{n-1}] m C(z)^{m-1} C'(z).$$
This  is by  the  Cauchy  Coefficient formula  ($C(z)$  analytic in  a
neighborhood of zero)
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} m C(z)^{m-1} C'(z) \; dz.$$
Now we put $C(z)= w$ so that $C'(z) \; dz = dw$ and use the functional
equation
$$\frac{m}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n}}{(w-1)^n} w^{m-1} \; dw
\\ = \frac{m}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n+m-1}}{(w-1)^n} \; dw
\\ = \frac{m}{2\pi i}
\int_{|w-1|=\gamma} \frac{1}{(w-1)^n}
\sum_{q=0}^{2n+m-1} {2n+m-1\choose q} (w-1)^q \; dw.$$
Therefore
$$\bbox[5px,border:2px solid #00A000]{
C_{n,m} = \frac{m}{n} \times {2n+m-1\choose n-1}.}$$
We can verify the alternate form
$${2n+m-1\choose n} - {2n+m-1\choose n-1}
\\ = \frac{n+m}{n} {2n+m-1\choose n-1} - {2n+m-1\choose n-1}
\\ = \frac{m}{n} \times {2n+m-1\choose n-1}.$$
Here the branch cut of $C(z)$  is $[1/4, \infty)$ (principal branch of
the logarithm). This is from Flajolet & Sedgwick, Analytic Combinatorics, page 732, Lagrange Inversion.
