Question

How to show that (kn)! is divisible by (k!)^n for all k,n >= 1?

Answer 1

I tried induction but I got nowhere XD

Answer 2

I have an idea. It is true that product of n consecutive integers is divisible by n!. Then (kn)! is k products of n consecutive integers so (kn)! is divisible by (k!)^n.

Answer 3

For example for n = 3 and k = 5 I get
(5!)^3 = (1 * 2 * 3 * 4 * 5)*(1*2*3*4*5)*(1*2*3*4*5).
(5*3)! = (15* 14 * 13 * 12 * 11)*(10*9*8*7*6)*(5*4*3*2*1).
then each group of 5 consecutive integers is divisible by 5! and there are k such grups so it is divisible by (5!)^3. Am I right?

Answer 4

$$
(kn)!=kn\times(kn-1)\times(kn-2)\cdot\cdot\cdot\times1\\
(k!)^n=\left (k\times(k-1)\times(k-2)\cdot\cdot\cdot\times1\right )^n\\
$$
so
$$
\cfrac{(kn)!}{(k)!^n}\\
=\cfrac{kn\times(kn-1)\times(kn-2)\cdot\cdot\cdot\times1}{\left (k\times(k-1)\times(k-2)\cdot\cdot\cdot\times1\right )^n}\\
$$
Let n be arbitray and let \(k\to\infty \):
$$
\lim_{k\to\infty}\cfrac{(kn)!}{(k!)^n}=\cfrac{(kn)^n}{k^n}=n^n<\infty\\
$$
Therefore, for any k,n>1, \((kn)!\) is divisible by \((k!)^n\)l