- #1

- 59

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## Main Question or Discussion Point

Hi, we know that for all interger [itex]m[/itex], [itex]n_1!\cdots n_k![/itex] divides [itex]m![/itex] where [itex]n_1+\cdots+n_k=m[/itex].

Now I want to show that [itex]m! (n!)^m[/itex] divides [itex](mn)![/itex].

We see that [itex](n!)^m[/itex] divides [itex](mn)![/itex] since [itex]\overbrace{n+\cdots+n}^{m-terms}=mn[/itex]

Also [itex]m!(nm-m)![/itex] divides [itex](mn)![/itex] similarly.

But how could I show my required statement above? It works for some examples that I've tested and I believe it is true.

Any inputs would be appreciated! :)

Now I want to show that [itex]m! (n!)^m[/itex] divides [itex](mn)![/itex].

We see that [itex](n!)^m[/itex] divides [itex](mn)![/itex] since [itex]\overbrace{n+\cdots+n}^{m-terms}=mn[/itex]

Also [itex]m!(nm-m)![/itex] divides [itex](mn)![/itex] similarly.

But how could I show my required statement above? It works for some examples that I've tested and I believe it is true.

Any inputs would be appreciated! :)