Relation between $(km)!$ and few others

Question

Show that $(m!)^k, (k!)^m, k!m!$ are all less equal $(km)!$ where $k$ and $m$ are integers greater than $1$.

I tried it by inducting on $k$ keeping $m$ fixed but got stuck at inductive step.

Answer 1

For any $0\leq j\leq k-1$, $$ \prod_{t=1}^m (jm+t)\geq \prod_{t=1}^m t=m! $$

So $$ (km)!=\prod_{j=0}^{k-1}\prod_{t=1}^m(jm+t)\geq \prod_{j=0}^{k-1} m!=(m!)^k $$

Switching the roles of $k$ and $m$ yields $(km)!\geq (k!)^m$.

For the last one, assume $k\leq m$ without loss of generality. If $k\geq 2$ then $$ (km)!\geq (m!)^k\geq m!m!\geq k!m!. $$ If $k=1$ then $(km)!=m!=k!m!$.

Added later: The first one can be done with a combinatorics argument instead. $(km)!$ is the number of ways of ordering $km$ objects. Split those $km$ objects into $k$ groups of $m$ each. Then the total number of orderings is at least the number of ways of ordering each of the $k$ groups of $m$ individually. The latter figure is $(m!)^k$ since each group has $m!$ orderings and there are $k$ groups.