${1\over{n+1}} {2n \choose n} = {2n \choose n} - {2n \choose n-1}$ I need to prove that $${1\over{n+1}} {2n \choose n} = {2n \choose n} - {2n \choose n-1}$$ 
I started by writing out all the terms using the formula ${n!\over{k!(n-k)!}}$ but I can't make the two sides equal. 
Thanks for any help.

That should work. Note that $$ \begin {align*} \dbinom {2n}{n-1} &= \dfrac {(2n)!}{(n-1)! (n+1)!} \\&= \dfrac {n}{n+2} \cdot \dfrac {(2n!)}{n!n!} \\&= \dfrac {n}{n+1} \cdot \dbinom {2n}{n}, \end {align*} $$ so we have: $ \dbinom {2n}{n-1} = \left( 1 - \dfrac {1}{n+1} \right) \cdot \dbinom {2n}{n}, $ which is equivalent to what we wanted to show. 
It seems I've been beaten by 2 answers . . .
