## 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.