How do I get the solution for a recursion with n! using Master Theorem?

Question

The problem in my book is T(n) = 16T(n/4) + n!

I don't know how to get at the correct solution my book has, which says Θ(n!)

here are pictures of the master theorem and the problem in my book.

Answer 1

Suppose for simplicity that we only consider $n = 4^k$, and that $T(1) = 1$.

Let us guess that $T(n) \leq Cn!$ (where $C>0$), and try to prove it by induction. For the inductive step we need $$ 16C(n/4)! + n! \leq Cn!, $$ that is, $$ 16C(n/4)! \leq (C-1)n!. $$ For the base case we need $C \geq 1$.

Let us write the inductive constraint for $n = 4$: $$ 16C \leq 24(C-1) \Longrightarrow C \geq 3. $$ This prompts us to choose $C = 3$. For $n \geq 4$ we then need $$ 48 (n/4)! \leq 2n! \Longleftrightarrow \frac{n!}{(n/4)!} \geq 24. $$ It is easy to check that this always holds – for $n = 16$ it's a simple computation, and for $n \geq 64$ we have $n!/(n/4)! \geq n > 24$.

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Intuitively, what is happening here is that $(n/4)!$ is much much smaller than $n!$, and so $T(n) \approx n!$, since all other summands (which you see if you unroll the recursion) are so much smaller.