An elegant and rather unexpected formula for the powers of the generating function for Catalan numbers $C_n = \frac{1}{n+1}\binom{2n}{n}$: $$\left(\sum_{n=0}^{\infty} \frac{1}{n+1}\binom{2n}{n} \cdot x^n\right)^m = \sum_{n=0}^{\infty} \frac{m}{n+m}\binom{2n+m-1}{n} \cdot x^n.$$ The formula can be further continued as $\ldots =\left(\frac{1-\sqrt{1-4x}}{2x}\right)^m$, but this would spoil the beauty of the above identity.
Max Alekseyev
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