Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$ 
Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$

Plugging in the Catalan Equation, we want to prove:
\begin{align*}
  \frac{1}{n+2} \binom{2(n+1)}{n+1} &= \sum\limits_{i=0}^n \frac{1}{i+1} \binom{2i}{i} \frac{1}{n-i+1} \binom{2(n-i)}{n-i} \\
\end{align*}
Expanding the binomial coefficients to factorials (again this is a formula we want to prove, not a result we have demonstrated):
\begin{align*}
  \frac{1}{n+2} \frac{(2n+2)!}{(n+1)!(n+1)!} &= \sum\limits_{i=0}^n \frac{1}{i+1} \frac{(2i)!}{i!i!} \frac{1}{n-i+1} \frac{(2n-2i)!}{(n-i)!(n-i)!} \\
\end{align*}
How would one go from here?

I think the best way to solve Segner's recurrence relation is to use generator functions.
