I am trying to solve the following problem:
Suppose that $f(z) = \sum_{k=0}^{\infty} c_kz^k$ is an analytic function in $\mathbb{D} = \{z\in\mathbb{C}:|z|<1\}$. Prove that $F(z)=\sum_{k=0}^{\infty} \frac{c_k}{k!}z^k$ is an entire function such that $|F(z)|\leq Me^{2|z|}$ in the whole complex plane for some constant $M$.
Progress: Since $f$ is analytic in $\mathbb{D}$, then the series $\sum_{k=0}^{\infty} c_kz^k$ converges for all $z\in\mathbb{D}$. In particular, $\sum_{k=0}^{\infty} \frac{c_k}{2^k}$ converges. It follows that the sequence $\left( \frac{c_k}{2^k}\right)_{k\geq 0}$ converges to $0$, so it is bounded. Let $M$ be a constant such that $\left|\frac{c_k}{2^k}\right|\leq M$ for all $k\in\mathbb{N}$. Hence, $$F(z) = \left| \sum_{k=0}^{\infty} \frac{c_k}{k!}z^k \right| \leq \sum_{k=0}^{\infty} \frac{|c_k|}{k!}|z|^k \leq M\sum_{k=0}^{\infty} \frac{(2|z|)^k}{k!} = Me^{2|z|}.$$
It is only missing to prove that $\sum_{k=0}^{\infty} \frac{c_k}{k!}z^k$ converges for all $z\in\mathbb{C}$.
By Cauchy-Hadamard, we have $$r=\limsup _{k\to\infty}\dfrac 1{\lvert \dfrac {c_k}{k!}\rvert ^{\frac 1k}}=\dfrac{\limsup_{k\to\infty}(k!)^{\frac1k}}{\limsup_{k\to\infty}\lvert c_k\rvert^{\frac1k}} =\dfrac {\infty}1=\infty.$$