Why is $\binom{n}{f}^g=O(n^{fg})$ true?
This reduces to showing $\binom{n}{f} = O(n^f)$.

In one variable the idea of an approximation being tight is pretty simple; you just see if the limit of the quotient goes to one.

In multiple variables there's no such thing as "the" limit.  There are limits along different curves.  Some may be tight and some may not be, but as long as the limit along one curve is tight there's no way of improving things across the board.

In situation at hand we have $$\binom{n}{f} = \frac{1}{f!}n(n-1)\ldots(n-f+1)$$.  For f small compared to n, the $(n-i) \approx n$.  If n grows a lot faster than k!, then it also blows out the k! term in the denominator as well.  So along a curve like this the bound is tight.

(It's also possible we're considering fixed k and just considering the growth as a function of n, in which case the $1/k!$ is just a constant factor.)