(n!)! is the product of the positive integers from 1 to n! we write integers from 1 to n! in (n – 1)! rows as follows

1.2.3......(n)

(n + 1)(n + 2) .... (2n)

(2n + 1)(2n + 2) .... (3n)

:

:

(n! – n + 1)(n! – n + 2) .... n(n – 1)!

Each of these (n – 1)! rows contain n consecutive positive integers. The product of consecutive integers in each row is divisible by n!

Hence to product of all integers from 1 to n! is divisible by (n!) (n–1)!

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