Show that, for every $n \in \Bbb N$, the following number is natural: $$\frac {(n!)!} {{n!}^{(n-1)!}}$$. I dont't know how to prove, as I tried to find a way including combinatorics.
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1Similar question: http://math.stackexchange.com/q/1684230/86846 – abiessu Apr 04 '16 at 18:19
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@abiessu How is this question similar to mine? – I. Stefan Apr 04 '16 at 18:21
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The methods used to prove the result will be similar. – abiessu Apr 04 '16 at 18:26
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See also this question, which implies the result. – Dietrich Burde Apr 04 '16 at 18:37
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$(n!)!$ is the product of $n! = n\cdot(n-1)!$ consecutive numbers. But if $a+1,a+2,\ldots,a+n$ are $n$ consecutive integers, then: $$ \frac{(a+1)(a+2)\cdot\ldots\cdot(a+n)}{n!} = \binom{a+n}{n}\in\mathbb{Z},$$ so your ratio is an integer, as the product of $(n-1)!$ integers.
Jack D'Aurizio
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Using the multinomial coefficient approach, we have
$$(n!)!\over n!\cdot n!\cdots n!$$
and the $\cdots$ represent the product over $(n-1)!$ total terms. Since each term is $n!$, we simply sum these indices and arrive at $n\sum_{i=0}^{(n-1)!}1=n!$, and we get the multinomial coefficient
$$\binom {n!}{n,n,n,\dots,n}$$
which is always an integer.
abiessu
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