Show that $(k!)^2$ divides $(2k+2)!$

Question

Show that $(k!)^2$ divides $(2k+2)!$

We have $\binom{2k+2}{k+2}=\dfrac{(2k+2)!}{k!(k+2)!}\in \Bbb Z$ say $=p$

Now $k!$ divides $(k+2)!\implies (k!)^2$ divides $(2k+2)!$.

Is the arguement correct?Please help.

Answer 1

Yes, your argument is correct and good.

Also, note that since $\binom{2n}{n}$ is an integer, we can similarly prove that $(n!)^2$ divides $(2n)!$.

More generally, it is a well-known result that $(n!)^k$ divides $(kn)!$.

Answer 2

We know that the product of any k consecutive integers is divisible by k!.

$(2k+2)!=[(2k+2).(2k+1)...(k+3)]\times[(k+2).(k+1)...3]\times 2.1$

where the first two factors in [] are each, products of k consecutive integers , so they are each divisible by k!, thus the whole number (2k+2)! is divisible by $(k!)^2$.