Show that $(k!)^n$ divides $(kn)!$

Question

Show that $(k!)^n$ divides $(kn)!$

I've tried it but without success. Any help would be great.

Answer 1

HINT:

$$\frac{(kn)!}{(k!)^n}=\binom{(kn)!}{\underbrace{k!,k!,\ldots,k!}_{n}}=\binom{(kn)!}{k!}\cdot\binom{(k(n-1))!}{k!}\cdot\binom{(k(n-2))!}{k!}\cdot\ldots\cdot\binom{k!}{k!}$$

is a multinomial coefficient that counts the ways to distribute $kn$ distinguishable objects into $n$ distinguishable boxes so that each box contains exactly $k$ objects.

Answer 2

HINT: It is enough to show, that every $2\leq j\leq k$ appears at least $n$ times in factors of $kn$.

Answer 3

Peace be upon you,

The nominator ($(kn)!$) has from each of the numbers $S=\{1,2,3,...k\}$, $n$ multiples. For example, for $i\in S$ it has $\{i,2i,3i,...,ni\}$.

Alongside, in denominator ($(k!)^n$), each of the numbers in $S$ has been repeated n times.

Obviously, the below one counts the above one.

Answer 4

$$(kn)! = (kn)(kn - 1)(kn - k + 1)...(k(n-1))...(k(n-1) - k + 1) ...(k)(k-1)...(1).$$ $$ = (kn)!/(k(n-1))! \times (k(n-1))!/(k(n-2))! \times ... \times k!$$

Each $(k(n-a))!/(k(n-a-1))!$ term is a multiple of numbers spanning k consecutive integers so it contains at least one multiple of each of $k, k-1, k-2, ..., 2$.