TITLE: Show that $(n!)^{(n-1)!}$ divides $(n!)!$
QUESTION [10 upvotes]: Show that $(n!)^{(n-1)!}$ divides $(n!)!$

I found this question in a text he was reading about BREAKDOWN OF PRIME FACTORS IN FACT, I decided many years, have posted some here to help, but this ... Did not come out at all, in fact I could not leave the place, and looked reolhei theorems that shows me the text, but could not solve, need help.

REPLY [2 votes]: This is the same as showing that combinations or binomial coefficients $\displaystyle{n\choose k}=C_n^k=\prod_{j=0}^{k-1}\frac{n-j}{1+j}$ are always natural, for all values of n and k. It all depends on proving that the product of any k consecutive numbers is always divisible through the product of the first k consecutive numbers, $1$ through k. This is obvious, since, in each sequence of $2$ consecutive numbers, exactly one is even, and one odd; in each sequence of three consecutive numbers, exactly one is ternary, while the other two aren't; in each sequence of four consecutive numbers, exactly one is quaternary, while the rest aren't; etc. Our product, $(n!)!=1\cdot2\cdot3\cdot\ldots\cdot n!$, can be broken up into $\displaystyle\frac{n!}n=(n-1)!$ sequences of n consecutive terms, each such sub-product being divisible through n!, for the reasons explained above. In other words, the whole product is divisible through $(n!)^{(n-1)!}$. QED.