${1\over{n+1}} {2n \choose n} = {2n \choose n} - {2n \choose n-1}$ 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.
 A: Because $(n+1)! = (n+1)n!$ and $n! = n(n-1)!$, we have
$$
\binom{2n}{n-1} = \frac{(2n)!}{(n-1)!(n+1)!} = \frac{n}{n+1}\binom{2n}{n}
$$
and the rest is easy.
A: $$ {2n \choose n}-{1\over{n+1}} {2n \choose n}$$
$$=\binom{2n}n\left(1-\frac1{n+1}\right)$$
$$=\frac{(2n)!}{n! n!}\cdot\frac n{n+1}$$
$$=\frac{2n!}{\{(n+1)\cdot n!\}\{n\cdot(n-1)!\}}\cdot n$$
$$=\frac{2n!}{(n+1)!\cdot (n-1)!}$$
$$ = {2n \choose n-1}= {2n \choose n+1}$$ 
A: 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 . . .
A: Multiply both sides by $n+1$. On RHS you get $\binom{2n}{n}(n-n+1)$ which will be equal to LHS. This is because 
$$
\binom{2n}{n-1}\cdot (n+1)=\binom{2n}{n-1}\cdot (n+1)\cdot \frac{n}{n}=n \cdot \binom{2n}{n}
$$
