Show that the derivatives of an analytic function cannot satisfy $|f^{(n)}(z)|>n!n^n$ for all $n$ for any $z$ where $f$ is analytic Show that the derivatives of an analytic function cannot satisfy $|f^{(n)}(z)|>n!n^n$ for all $n$ for any $z$ where $f$ is analytic.
My attempt: 
Assume otherwise.
Say $|f^{(n)}(z)|>n!n^n$.
$a_n$, the $n$th term of the Taylor series for $f$ centered at $z$ is $a_n=\frac{f^{(n)}(z)}{n!}$ and $|a_n|\leq\frac{M}{r^n}$.
Thus we have 
$$n^n<\bigg|\frac{f^{(n)}(z)}{n!}\bigg|=|a_n|\leq\frac{M}{r^n}$$
$$\Rightarrow n^n<\frac{M}{r^n}$$
Which is clearly false for large $n$ but not necessarily for all $n$.
I also tried this:
Assume otherwise.
Say $|f^{(n)}(z)|>n!n^n$.
Since $f(z)$ is analytic, by Cauchy's theorem
$$\bigg|f^{(n)}(z)\bigg|=\bigg|\frac{n!}{2\pi i}\int_{|w-z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|$$
$$\Rightarrow\frac{1}{2\pi}\bigg|\int_{|w-z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|>n^n$$
but again I don't know where to go with this.

By the post Find the upper bound of the derivative of an analytic function, if $|f(z)|\leqslant M$ on $|z|\leqslant r$, then
$$
\bigg|f^{(n)}(z)\bigg|\leqslant \frac{M n!\rho^n}{(r-|z|)^{n}}
$$ 
where $\rho>1$. If $|f^{(n)}(z)|>n!n^n$ for all $n$ for any $z$, let $z=0$ and we have
$$
n!\cdot n^n<\frac{M n!\rho^n}{(r-|z|)^{n}}\quad\text{or }\quad (\frac{rn}{\rho})^n<M 
$$
This is impossible since left is unbounded for large $r$ and $n$. 
