Can an entire function with exponential growth have bounded derivatives?

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Here is this week's POTW:

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Let $F$ be an entire function for which there exists $t > 0$ such that $\lvert F(z)\rvert = O(\exp(\lvert z\rvert^t))$ as $\lvert z\rvert \to \infty$. Show that there is a constant $M > 0$ such that for all $n$ sufficiently large, $$\lvert F^{(n)}(0)\rvert \le Mn!\left(\frac{et}{n}\right)^{n/t}$$

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This week's problem was solved correctly by GJA. Here is his solution.

Spoiler

Since $F$ is entire, we can apply Cauchy's inequalities to disks of arbitrary radius centered about the origin to obtain

$\left|F^{(n)}(0)\right|\leq \dfrac{n!\|F\|_{C}}{R^{n}},$

where $\|F\|_{C}=\sup_{z\in C}|F(z)|$ denotes the supremum of $|F|$ on the boundary circle - $C$ - of the disk centered at the origin with radius $R$.

From the order condition on $f$, there is $N\in\mathbb{N}$ such that $|F(z)|\leq M\exp{\{|z|^{t}\}}$ for all $|z|\geq \left(\dfrac{N}{t}\right)^{1/t}$.

Now fix $n\geq N,$ set $R=\left(\dfrac{n}{t}\right)^{1/t},$ and consider the disk centered about the origin of radius $R$. Applying the above considerations, we have

\begin{align*}
\left|F^{(n)}(0)\right|&\leq \dfrac{n!\|F\|_{C}}{R^{n}}\\
&\leq Mn!\left(\dfrac{t}{n}\right)^{n/t}\|\exp\{|z|^{t}\}\|_{C}\\
&=Mn!\left(\dfrac{t}{n}\right)^{n/t}e^{n/t}\\
&=Mn!\left(\dfrac{et}{n}\right)^{n/t},
\end{align*}

thereby establishing the result for all sufficiently large $n.$