Showing that $|f^{(n)}| \le n!n^n$ and then making this result sharper

Solution 1:

The inequality you derive in the 4th point $$ \lvert f^{(n)}(z) \rvert \le \frac{n! M}{r^n} $$ is enough to substantially improve the result:

if $g(n)$ is any increasing function that increases to $\infty$, no matter how slowly, then chose $n_0$ such that $f(z)$ is analytic in the circle centered in $z$ of radius $$ \frac{2}{g(n_0)} $$ and such that $2^{n_0} > M$ then for any $n>n_0$ using your inequality in the circle of radius $$ r=\frac{2}{g(n)} < \frac{2}{g(n_0)}$$ gives $$ \lvert f^{(n)}(z) \rvert \le \frac{n!M}{r^n} =n!M2^{-n}g(n)^n \leq n! 2^{n_0} 2^{-n}g(n)^n \leq n! 2^{n} 2^{-n} g(n)^n =g(n)^nn!. $$

This result is essentially best possible, to see it, suppose that the inequality $$ \lvert f^{(n)}(z) \rvert \le H(n)^n n! $$ is true for certain bounded increasing function $H(n)$ and for all the analytic functions $f(z)$, then if $C > H(n)$ for all $n$ then consider the funcion $$ h(z) = \sum_{n=0}^\infty (2C)^n z^n $$ this function is analytic inside the circle of radius $1/2C$, but we have $$ \lvert h^{(n)}(0) \rvert = (2C)^n n! > H(n)^n n! $$ a contradiction.