Question: Let $f$ be an analytic function in the unit disc $D={z\in C: |z|<1}$. Consider a point $z_0\in D$. Show that there must be a positive integer n such that the n-th derivative of $f$ at $z_0$ satisfies $|f^{(n)}(z_0)| \leq n!n^n$.
Thoughts so far: In class, we discussed power series and their usefulness in proving a variety of facts of complex analysis. Just by rearranging the inequality we see that $\frac{|f^{(n)}(z_0)|}{n!} \leq n^n$. Clearly $\frac{|f^{(n)}(z_0)|}{n!}$ is a term of the Taylor series expansion for $f(z)$, but the $n^n$ term is annoying the royal bananas out of me as I cannot think of a way to get that. I am considering using the ratio ($\lim_{n\to\infty} \frac{\frac{f^{(n)}(z_0)}{n!}}{\frac{f^{(n-1)}(z_0)}{(n-1)!}} = \lim_{n\to \infty} \frac{f^{(n)}(z_0)}{nf^{n-1}(z_0)} \leq 1 \implies \lim_{n\to\infty} \frac{f^{(n)}(z_0)}{f^{n-1}(z_0)} \leq n$) or n root test ($\lim_{n\to\infty} \sqrt[n]{\frac{f^{(n)}(z_0)}{n!}} \leq 1$) for convergence since analyticity implies convergence of the series, but I cannot seem to coax the $n^n$ out of any of my attempts. Also, I am uncertain whether I am using the notion of analyticity and power series correctly.
Thank you in advance for any help that you may provide.