\(\displaystyle n(!)^k \) is divisible by \(\displaystyle n!\)
Now just prove that \(\displaystyle n(!)^k > (n!)^{(n(!)^0 -1)(n(!)^1 -1)...(n(!)^{k-2}-1)}\)

The two sides have the same elements(factors).
n(!)^k has more factors , same as being greater than the left side.
Proving that it can be divided by the left side is same as proving it is greater.