Showing that $(k!)!$ is divisible by ${(k!)}^{(k-1)!}$ using a combinatoric argument?

I know that $k!$ is the number of ways to order $k$ distinct objects where order matters

So then $(k!)!$ is the number of ways to order $k!$ distinct objects where order matters.

So then $(k!)^{(k-1)!}$ means we have $k$ of $(k-1)!$ objects of the same kind that each have $k!$ ways of being ordered where order matters

The argument is how many ways are there to permute $(k!)!$ objects with $(k-1)!$ distinct objects (number of distinct objects being $k!$, i.e. $k!$ different color balls with $(k-1)!$ of each color ball) such that the order of these non-distinct objects don't matter?

And that is $$\frac{(k!)!}{(k!)^{(k-1)!}}$$

I am confused if I am on the right track here, because this seems very abstract / confusing to me.

HINT: As usual, for any positive integer $m$ let $[m]=\{1,\ldots,m\}$. Show that for any $k,n\in\Bbb Z^+$,
$$\binom{kn}{\underbrace{k,k,\ldots,k}_{n\text{ copies}}}=\frac{(kn)!}{(k!)^n}\;,$$
is the number of ordered partitions $\langle A_1,\ldots,A_n\rangle$ of $[kn]$ into $n$ $k$-element sets, and apply this to the case $n=(k-1)!$.
$$\binom{3k}k\binom{2k}k=\frac{(3k)!}{k!(2k)!}\cdot\frac{(2k)!}{k!\cdot k!}=\frac{(3k)!}{(k!)^3}$$
ordered partitions of $[3k]$ into $3$ $k$-element sets.