So the nth catalan numbers is given by $$C_n = \frac{1}{n+1} {2n \choose n}$$
I want to prove now that $n+1$ divides ${2n \choose n}$ so I try to do it the following way $${2n \choose n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!^2} = \frac{(2n)(2n-1)!}{(n(n-1)!)^2}$$
But still How can I conclude that $n+1$ divides ${2n \choose n}$ ?
