Prove that there does not exist a holomorphic function f(z) on any open set containing 0 such that $f^{(n)}(0) = n^n\cdot n!$
I tried to use the Cauchy integral formula for higher derivatives and proved by contradiction, but had no progress. Could someone give some insights on this?
Thank you very much.
Hint: Suppose there were such an $f.$ Then
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n $$
has positive radius of convergence.