You can probably find it somewhere online, but for completeness here’s a derivation of the familiar closed form for $C_n$ from the recurrence $$C_n=\sum_{k=0}^{n-1}C_kC_{n-1-k}\tag{0}$$ and the initial value $C_0$, via the ordinary generating function. Then, as in Mhenni Benghorbal’s answer, you can easily (discover and) verify the first-order recurrence. I don’t see any nice way to get it directly from $(0)$.
Let the ordinary generating function for the Catalan numbers be
$$c(x)=\sum_{n\ge 0}C_nx^n=\sum_{n\ge 0}\sum_{k=0}^{n-1}C_kC_{n-1-k}x^n\;.$$
Then since $C_0=1$, we have
$$\begin{align*}
c(x)&=\sum_{n\ge 0}\sum_{k=0}^{n-1}C_kC_{n-1-k}x^n\\
&=1+\sum_{n\ge 1}\sum_{k=0}^{n-1}C_kC_{n-1-k}x^n\\
&=1+x\sum_{n\ge 0}\sum_{k=0}^nC_kC_{n-k}x^n\\
&=1+x\left(\sum_{n\ge 0}C_nx^n\right)^2\\
&=1+xc(x)^2\;,
\end{align*}$$
or $xc(x)^2-c(x)+1=0$. The quadratic formula then yields
$$c(x)=\frac{1\pm\sqrt{1-4x}}{2x}\;,\tag{1}$$
and since
$$\lim_{x\to 0}c(x)=\lim_{x\to 0}\sum_{n\ge 0}C_nx^n=C_0=1\;,$$
it’s clear that we must choose the negative square root in $(1)$, so that
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}\;.$$
Now apply the binomial theorem to $\sqrt{1-4x}$:
$$\begin{align*}
\left(1-4x\right)^{1/2}&=1+\sum_{n\ge 1}\binom{1/2}n(-4x)^n\\
&=1+\sum_{n\ge 1}\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\dots\left(-\frac{2n-3}2\right)}{n!}(-4x)^n\\
&=1+\sum_{n\ge 1}(-1)^{n-1}\frac{(2n-3)!!}{2^nn!}(-4x)^n\\
&=1-\sum_{n\ge 1}\frac{2^n(2n-3)!!}{n!}x^n\\
&=1-2\sum_{n\ge 1}\frac{2^{n-1}\prod_{k=1}^{n-1}(2k-1)}{n(n-1)!}x^n\\
&=1-2\sum_{n\ge 1}\frac{2^{n-1}(n-1)!\prod_{k=1}^{n-1}(2k-1)}{n(n-1)!^2}x^n\\
&=1-2\sum_{n\ge 1}\frac{\left(\prod_{k=1}^{n-1}(2k)\right)\left(\prod_{k=1}^{n-1}(2k-1)\right)}{n(n-1)!^2}x^n\\
&=1-2\sum_{n\ge 1}\frac{(2n-2)!}{n(n-1)!^2}x^n\\
&=1-2\sum_{n\ge 1}\frac1n\binom{2(n-1)}{n-1}x^n\;,
\end{align*}$$
so
$$\begin{align*}
c(x)&=\frac1{2x}\cdot2\sum_{n\ge 1}\frac1{n}\binom{2(n-1)}{n-1}x^n\\
&=\sum_{n\ge 1}\frac1n\binom{2(n-1)}{n-1}x^{n-1}\\
&=\sum_{n\ge 0}\frac1{n+1}\binom{2n}nx^n\;,
\end{align*}$$
and we have the familiar closed form $C_n=\dfrac1{n+1}\dbinom{2n}n$.