Showing that $|f^{(n)}| \le n!n^n$ and then making this result sharper 
Ahlfors: Show that the successive derivatives of an analytic function at a
  point can never satisfy $|f^{(n)}(z)| > n!n^n$.  Formulate a sharper
  theorem of the same kind.

Attempt for Part One:


*

*Let $\Delta$ be a neighborhood around $z$ of radius $r$ small enough so that $f$ is analytic on $\Delta$.  Let $C$ be a circle around $z$ of radius $r$.

*Cauchy's integral formula then yields that
$$
f^{(n)}(z) = {n! \over 2 \pi i}\int_{C} {f(\zeta) \over (\zeta - z)^{n+1}}\ d\zeta
$$

*Since $\mathbb{R}$ is complete, we have that $M = \max\{|f(\zeta)| : |\zeta - z| \le r \} \in \mathbb{R}$ exists.  Furthermore, we have that $|\zeta - z| \le r$ for all $\zeta$ within the perimeter of $C$.  Hence we have
$$
|f^{(n)}(z)| \le \left|{n! \over 2 \pi i}\right|\int_{C} {|M| \over |r^{n+1}|}\ |d\zeta| \le {n! \over 2 \pi} {M 2 \pi r \over r^{n+1}} = {n!M \over r^{n}}
$$

*Hence we have Cauchy's estimate:
$$
|f^{(n)}(z)| \le {n! M \over r^n}
$$

*We may further assume that $M > 1$ above (it doesn't affect any of the the inequality reasoning).

*Consider that there is a point where $n$ is large enough s.t. $n^n\ge {M \over r^n}$ (indeed, it is when $n \ge {M \over r}$).  At such a point, we have that:
$$
n^n \ge {M \over r^n} \implies n^n r^n \ge M \implies r^n \ge {M \over n^n} \implies {1 \over r^n} \le {n^n \over M}
$$

*Then assuming $n^n \ge {M \over r^n}$, we have that
$$
\underbrace{|f^{(n)}(z)| \le {n!M \over r^n}}_{\text{Cauchy's estimate}} = {n!M} \cdot \left({1 \over r^n}\right) \le \underbrace{{n!M}\cdot \left({n^n \over M}\right)}_{\text{since }{1 \over r^n} \le {n^n \over M}} = n! n^n
$$
as desired.

*We have thus far shown that if $n \ge {M \over r}$, then 
$$
|f^{(n)}(z)| \le n!n^n
$$
Question: In what way could we use this result to make a sharper theorem of the same kind?
 A: 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.
