holomorphic function on punctured disk satisfying $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ has an essential singularity at $0$

Let $f$ be holomorphic non-constant on $D=\left\{ 0<\left|z\right|<10\right\}$ . Given that for all $n\in\mathbb{N}$: $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ prove that $f$ has an essential singularity at $0$. Find an example of $f$ satisfying the condition.

My idea was to assume for contradiction the singularity is not essential, which means that either the limit $z\to0$ of $f$ exists or of $1/f$, and from $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ we get that the limit of $f$ exists and is $0$.

I then define $h(z)=f(z)$ for $z\neq0$ and $h(0)=0$, which is holomorphic on the entire disk, and derive a contradiction from there and the uniqueness theorem. But then the suitable holomorphic function is the Gamma function, which we have not really discussed in class and I’m not sure about it’s properties.

Any ways of solving this which avoids using the gamma function?

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Any ways of solving this which avoids using the gamma function?

Let $g$ be a non-constant entire function with zeros at all integers, and look at

$$f(z) = g(1/z).$$

Then $f$ is holomorphic on $\mathbb{C}\setminus \{0\}$, so surely on the punctured disk $0 < \lvert z\rvert < 10$, and we have

$$\biggl\lvert f\biggl(\frac{1}{n}\biggr)\biggr\rvert = 0 < \frac{1}{n!}$$

for all $n$.

Clearly, $f$ cannot have a pole at $0$, since that would imply $\lvert f(z_n)\rvert \to +\infty$ for all sequences $(z_n)$ in the punctured disk converging to $0$. To exclude a removable singularity at $0$, one shows that if $h \colon \{ z : \lvert z\rvert < 10\} \to \mathbb{C}$ is holomorphic with $\bigl\lvert h\bigl(\frac{1}{n}\bigr)\bigr\rvert \leqslant \frac{1}{n!}$ for all $n\in \mathbb{N}$, then $h \equiv 0$. For that, use the power series expansion

$$h(z) = \sum_{k = 0}^{\infty} a_k z^k$$

and show that if there is an $m$ with $a_m \neq 0$, the inequality $\bigl\lvert h\bigl(\frac{1}{n}\bigr)\bigr\rvert \leqslant \frac{1}{n!}$ cannot hold for large enough $n$.