Prove that $(n!)!$ is divisible by $(n!)^{(n - 1)!}$ .

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Solution

Now, $(n!)!$ is the product of the positive integers from $1$ to $n!$ .
We write the integers from $1$ to $n!$ in $(n - 1)!$ rows as follows:
$1 \times 2 \times 3 . . . n$
$(n + 1) (n + 2) (n + 3) . . . (2 n)$
$(2 n + 1) (2 n + 2) . . . (3 n)$
$(3 n + 1) (3 n + 2) . . . (4 n)$
.
.
.
$(n! - n + 1) (n! - n + 2) . . . (n (n - 1)!)$
Each of these $(n - 1)!$ rows contain $n$ consecutive positive integers. The product of consecutive integers in each row is divisible by $n!$ Thus, the product of all the integers from $1$ to $n!$ is divisible by $(n!)^{(n - 1)!}$ .

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