let $$a_n^{<m>} \equiv \sum_{k_1+k_2+...+k_m = n}a_{k_1}a_{k_2}...a_{k_m}\quad k_i\in \mathbb{Z}^+\cup\{0\} $$ and $$C_n = \frac{1}{n+1}\binom{2n}{n}$$ (note that $C_{n+1}=C_n^{<2>}$)
prove for $n\geq3$ $$C_n = \sum_{m = 1}^{n}C_{n-m}^{<m>}$$ he also gave a hint: remember that $C_n$ is the number of "good walk" from $(0,0)$ to $(n,n)$ i try induction but got nowhere