When $n$ is a natural number, prove that $\frac{(2n)!}{n!^2}$ is a natural number.

If possible, I would like you to prove this by induction.

I tried to prove this by induction, but I can’t because $k+1$ is left in the denominator when I substitute $k+1$ to $n$.

Help me to solve this.


Doesn't this suffice? $$ \frac{(2n)!}{n!^2} =\frac{(2n)!}{n!\,n!} =\binom{2n}{n} $$

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  • $\begingroup$ Still you have to prove that it is an integer: math.stackexchange.com/questions/382787/… $\endgroup$ – Guy Fsone Dec 6 '17 at 12:14
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    $\begingroup$ @GuyFsone, not if you use that $\binom{m}{n}$ is the number of subsets of size $n$ out of $m$ objects. $\endgroup$ – lhf Dec 6 '17 at 12:17

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