My original goal was to show that, if $f$ is analytic on the unit disk and $|f(z)|\leq \frac 1 {1-|z|}$, then $|f^{(n)}(0)|\leq e(n+1)!$
I have managed to show that, given $0<R<1$, we can use Cauchy's estimate to obtain $|f^{(n)}(0)| \leq n!R^{-n}\frac 1 {1-|R|}= \frac {n!} {R^n-R^{n+1}}$. Now, this is $\leq e(n+1)!$ when $\frac 1 {e(n+1)} \leq R^n-R^{n+1}$.
So, if I can show that $\frac 1 {e(n+1)} \leq R^n-R^{n+1}$ for at least one $0<R<1$, I'll be done.
This is obviously true when $n=0$, but I don't know how to show the general case.