Prove that the product of 2n consecutive negative integers is divisible by (2n)!

Let $2 n$ negative integers be $(-r),(-r-1),(-r-2), \ldots, \ldots,(-r-2 n+1)$.

Then, product $=(-1)^{2 n}(r)(r+1)(r+2), \ldots \ldots, \ldots(r+2 n-1)$

$=\frac{(r-1) !(r)(r+1)(r+2) \ldots \ldots(r+2 n-1)}{(r-1) !}$

$=\frac{(r+2 n-1) !}{(r-1) !}$

$=\frac{(r+2 n-1) !}{(r-1) !(2 n) !} \times(2 n) !$

$={ }^{r+2 n-1} C_{2 n} \times(2 n) !$

This is divisible by $(2 n) !$.