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Consider the question: does there exist a holomorphic function $f$ on the unit disk in the complex plane such that $f\left({1 \over n}\right) = e^{-n}$ ? I came up with an answer but I’d like to know if there is another argument.

Here’s my answer: let $f$ be such a function. $f$ must vanish at $0$ by continuity. Any holomorphic function vanishes at $0$ satisfies the following estimate : $\left| f(z) \right|\sim_{z \to 0} \alpha \,|z|^m$, where $\alpha>0$ and $m$ is a positive integer (this is easily derived from the expression of $f$ as a power series). This contradicts the assumption $f\left({1 \over n}\right) = e^{-n}$.

By curiosity, I would like to know if there is another argument (maybe something about $f(z) – e^{1 \over z}$)?

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Your argument is correct. Expressed more formally, it could go as follows: suppose such $f$ exists. Let $k\ge 0$ be the smallest integer such that $f^{(k)}(0)\ne 0$. Then the limit $\lim_{z\to 0}z^{-k}f(z)$ exists and is not zero. But plugging $z=1/n$ here gives $\lim_{n\to\infty } n^k e^{-n} $ which is zero.

Looking at the difference $g(z)=f(z)-e^{1/z}$ is not as helpful because $g$ is not holomorphic at $0$. So, the fact that the zeros of $g$ accumulate at $0$ is not a contradiction.

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