Is (nk)!/(n!)^k an Integer?

\[\frac{80!}{(20!)^4}\]

a whole number?
Yes. To see why. pick a prime number, say 19.

\[80!\]

includes as a factor

\[19 \times 38 \times 57 n\times 76=4! \times 19^4\]

and the denominator includes the factor

\[19!\]

. The first factor in the numerator is obviously an integer on division by the second factor, in the denominator. The same argument is true for any prime that os a factor of

\[20\]

.
We could also write

\[\frac{80!}{(20!)^4}= (\frac{80!/60!}{20!}) \times (\frac{60!/40!}{20!}) \times (\frac{40!/20!}{20!}) \times (\frac{10!/1!}{20!})\]

.
Each of these factors is an integer, so

\[\frac{80!}{(20!)^4}\]

is also an integer. In fact,

\[\frac{(nk)!}{(n!)^k}\]

is an integer in general.