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Can $(k!)!$ be shown to be divisible by $(k!)^{(k-1)!}$? What would be the simplest way of showing this?

M. Berkham
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2 Answers2

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Lemma: $a!$ divides the product of any $a$ consecutive integers.

Notice that $(k!)!$ is the product of the products of $\frac{k!}{k}=(k-1)!$ blocks of $k$ elements each ( the first block is $1\times 2\times \dots \times k$) , so by the lemma we have that $k!^{(k-1)!}$ divides $(k!)!$

Asinomás
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    In fact, the quotient is the multinomial coefficient $$ \frac{(k!)!}{k!^{(k-1)!}} = \binom{k!}{\underbrace{k,k,\cdots,k,k}_{(k-1)!}}. $$ – anon May 30 '17 at 01:59
  • Oh yeah, my apologies. It should be ok now. – Asinomás May 30 '17 at 02:03
  • The product of the $a$ consecutive integers is equal to $a!$, right? – M. Berkham May 30 '17 at 02:56
  • @M.Berkham not always, for example $3\times 4\times 5\neq 3!$. – Asinomás May 30 '17 at 03:06
  • Oh, I see. Is there a specific name for that lemma? It seems very useful. – M. Berkham May 30 '17 at 03:08
  • I dont think so, but you can find a prrof here https://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial. I also answered a similar question a while back here https://math.stackexchange.com/questions/1311466/prove-that-the-expression-frac3n3n-is-integral-for-n-geq-0/1311477#1311477 – Asinomás May 30 '17 at 03:12
  • Thanks. Is the multinomial coefficient always an integer? – M. Berkham May 30 '17 at 03:13
  • yes, at least when the numbers inside the bracket are positive integers and the numbers in the bottom add the same as the number in the top. That is because the multinomial coefficient has a combinatorial interpretation, so it counts things. – Asinomás May 30 '17 at 03:20
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    Got it. Thanks. – M. Berkham May 30 '17 at 03:22
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In fact, the quotient is the multinomial coefficient $$ \frac{(k!)!}{k!^{(k-1)!}} = \binom{k!}{\underbrace{k,k,\cdots,k,k}_{(k-1)!}}. $$

One can go considerably beyond this:

If $(c_1,c_2,\cdots,c_n)$ is a nonnegative integer solution to $1c_1+2c_2+\cdots+nc_n$ then

$$ \frac{n!}{1!^{c_1}c_1!~2!^{c_2}c_2!~\cdots~ n!^{c_n}c_n!} $$

counts the conjugacy class of $S_n$ consisting of all permutations with cycle type $1^{c_{\large 1}}2^{c_{\large 2}}\cdots n^{c_{\large n}}$ (which is shorthand for having $c_r$-many $r$-cycles for $r=1,\cdots,n$). This follows from the orbit-stabilizer theorem: $S_n$ acts transitively on conjugacy classes and the stabilizer of a permutation with said cycle type is an internal direct product of internal wreath products.

Specialize to the case $n=k!$ and $c_r=(k-1)!$ when $r=k$ and $c_r=0$ otherwise, giving

$$ \frac{ (k!)!}{ ((k-1)!)!~ k!^{(k-1)!}} $$

is an integer.

anon
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