Let $f: \Bbb{C} \to \Bbb{C}$ be an analytic function in $\Bbb{C}$ for which we know $|f^{(n)}(0)| \leq 1, \forall n\in\Bbb{N}$. Prove that, for any $z \in \Bbb{C}$, we have $$|f(z)| \leq e^{|z|}$$
Having that information about all derivatives of $f$ in conjunction with having an exponential function in the inequality I want to prove, got me thinking I could follow this path:
Given that $f$ is analytic in $\Bbb{C}$, $f$ has a Laurent series, centered at $0$, that converges for any $z$. Thus we have
$$f(z) = \sum_{n=0}^{\infty} c_nz^n$$
Furthermore, we know that
$$c_n = \frac{f^{(n)}(0)}{n!} \implies |c_n| \leq \frac{1}{n!}$$
With this in mind we can proceed as follows:
$$|f(z)| = \left|\sum_{n=0}^{\infty} c_nz^n\right| \leq \sum_{n=0}^{\infty} |c_nz^n| = \\ = \sum_{n=0}^{\infty} |c_n||z|^n \leq \sum_{n=0}^{\infty} \frac{|z|^n}{n!} = e^{|z|}$$
which concludes the proof. I am not exactly sure, however, that my starting point is right. That is, I do not know if I can assume from the get-go that there is a Laurent series with convergence radius $+\infty$.
I would appreciate alternative approaches to this problem - if anyone finds any - and considerations about the correctness of mine.