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

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