Prove that $\frac{(2n)!}{(n!)^2}-1$ is divisible by $(2n+1)$ 
Prove that $$\frac{(2n)!}{(n!)^2}-1$$ is divisible by $(2n+1)\;,$ Where $n\in \mathbb{N}$ and $n>1$

$\bf{My\; Try::}$  Let $$S = \frac{(2n)!}{n!^2}-1 = \frac{2^n(2n-1)(2n-3)\cdot \cdot ........\cdot 3 \cdot 2 \cdot 1}{n!}-1$$
Now How can we prove that $2^n(2n-1)(2n-3)\cdot \cdot \cdot .... 3\cdot 2 \cdot 1$ is divisible by $n!$ and odd number
Help Required, Thanks

$\dfrac{(2n)!}{(n!)^2} - 1 = \dbinom{2n}{n}-1$
This is the count of ways to arrange $n$ discrete objects into two equal piles, excluding a particular arrangement.
$\begin{array}{l:l:l} n & \binom{2n}n-1 & 2n+1
\\ 1 & 1 & 3
\\ 2 & 5 & 5
\\ 3 & 19 & 7
\\ 4 & 69 & 9
\\ \vdots & \vdots & \vdots
\end{array}$
Only occasionally is $\binom{2n}n -1$ divisible by $2n+1$
