Prove factorial problem $\forall n\in\mathbb N$ $C_n=\frac{1}{n+1}\left(\begin{matrix}2n\\n\end{matrix}\right)$. 
Prove that $C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)$ 
I tried many ways but got nothing. Appreciate any tips.
My attempt: By defining, $C_{n+1}= \dfrac{1}{n+2}\left(\begin{matrix}2n+2\\n+1\end{matrix}\right)=\dfrac{1}{n+2}\dfrac{(2n+2)!}{(n+1)!(n+1)!}=\dfrac{1}{n+2}*\dfrac{(2n+2)(2n+1)n!}{(n+1)^2n!n!}$
$=\dfrac{1}{n+2}\dfrac{2(2n+1)n!}{(n+1)n!}$
 A: You’ve misstated the result: you want to show that
$$C_n=\binom{2n}n-\binom{2n}{n-1}\;,$$
not that
$$C_{n+1}=\binom{2n}n-\binom{2n}{n-1}\;.$$
Note that $\binom{2n-1}{n-1}=\binom{2n-1}n$, so
$$\binom{2n}{n-1}=\frac{2n}{n+1}\binom{2n-1}{n-1}=\frac{n}{n+1}\left(\binom{2n-1}{n-1}+\binom{2n-1}n\right)\;.$$
Now use Pascal’s identity and a little algebra.
A: You need to simplify the second term: $\binom{2n}{n-1}=\dfrac{(2n)!}{(n-1)!(n+1)!}= \dfrac{(2n)!}{\frac{n!}{n}\cdot (n!\cdot (n+1))}= \dfrac{n}{n+1}\cdot \binom{2n}{n} \Rightarrow C_n = \binom{2n}{n} - \dfrac{n}{n+1}\binom{2n}{n}= \left(1-\dfrac{n}{n+1}\right)\binom{2n}{n} = \dfrac{1}{n+1}\binom{2n}{n}$.
A: $$C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)=\frac{1}{n+1}[\frac{(2n)!(n+1))}{n!n!}-\frac{(2n)!}{n!(n-1)!}]=\\\frac{1}{n+1}\frac{2n!}{n!}[\frac{n+1}{n!}-\frac{1}{(n-1)!}]=\\ \frac{1}{n+1}\frac{2n!}{n!n!}(n+1-\frac{n!}{(n-1)!})=\\ \frac{1}{n+1}\frac{2n!}{n!n!}(n+1-n)=\frac{1}{n+1}\binom{2n}{n}$$
