I am trying to solve this recurrence using the Master Theorem, but I am struggling to find a constant that satisfies the regularity condition.
For $T(n) = 16T(n/4) + n!$, we have $a = 16$, $b = 4$, $f(n) = n!$, and thus we have that $n^{\log_b{a}} = n^{\log_{4}{16}} = n^{2} = \Theta(n^2)$. $f(n) = \Omega(n^{\log_{4}{16} + \epsilon})$, for any $\epsilon > 0$, since $n!$ grows asymptotically faster than any polynomial.
So for this recurrence, Case 3 of the Master Theorem applies if we can show that the regularity condition holds for $f(n)$. That is, $a \cdot f(n/b) = 16 \cdot (n/4)! \leq c \cdot n! = c \cdot f(n)$ for all sufficiently large $n$.
At this point, how do I show that there exists a constant, $c < 1$, that satisfies the condition?
My hunch is that all $c \in (0, 1)$ should satisfy the condition. When considering the asymptotic behavior of $(n/4)!$ and $n!$, it seems that the constants will be insignificant for very large $n$, since $(n/4)! \ll n!$.