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 $n$.