Solve the recursion $$a_n=\sum^{n-1}_{k=0}a_{k}a_{n-k-1}=a_0a_{n-1}+a_1a_{n-2}+\cdots+a_{n-1}a_0$$
where $a_0=a_1=1$.
Let $$f(x) = a_0x + a_1x^2+a_2x^3 + \cdots + a_nx^{n+1}+\cdots$$
Then the desired result, $a_n$, is the coefficient of the term $x^{n+1}$. $$\begin{align*} f^2(x)=\ & a_0^2x^2 + (a_0a_1+a_1a_0)x^3 + \cdots+\sum_{k=0}^{n-1}a_ka_{n-k-1}x^n +\cdots\\=\ & a_1 x^2 + a_2x^3 + \cdots + a_{n}x^{n+1} + \cdots\quad(\because a_0=a_1=1)\\=\ &f(x) - x \end{align*}$$
Solving this equation and also note that $f(0)=0$ yields $$f(x)=\frac{1-\sqrt{1-4x}}{2}$$
The conclusion from the previous practice states that $$\sqrt{1+x}=1+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\cdot 2^{2n-1}}\binom{2n-2}{n-1}x^n$$
Replacing $x$ with $-4x$ gives $$\sqrt{1-4x}=1-\sum_{n=1}^{\infty}\frac{2}{n}\binom{2n-2}{n-1}x^n$$
Therefore, $$f(x)=\sum_{n=1}^{\infty}\frac{1}{n}\binom{2(n-1)}{n-1}x^n$$
By the definition of $f(x)$, $a_n$ is the coefficient of the term $x^{n+1}$ which means that $$a_n=\boxed{\frac{1}{n+1}\binom{2n}{n}}$$
which is a Catalan number.