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Prove that ((n^(2))!)/((n!)^(n)) is a na...

Prove that `((n^(2))!)/((n!)^(n))` is a natural number for all n `in` N.

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We have to prove that `(n^(2))!` is divisible by `(n!)^(n)`.
We know that product of r consecutive integers is divisible by r!
Now `(n^(2))!=1xx2xx3xx4xx..xx n^(2)`
`=(1xx2xx3xx..xx n)xx`
`[(n+1)(n+2)xx.. .. Xx(2n)]xx`
`[(2n+1)(2n+2).. .. (3n)]xx`
.. ..
.. ..
`[(n^(2)-(n-1)(n^(2)-n).. .. n^(2)]`
Thus `(n^(2))!` consists n groups of product of n consecutive integers.
Each group is divisible by n!.
So, `(n^(2))!` is divisible by `(n!)^(n)`.
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