Non-existence of a certain holomorphic function on the unit disk I am trying to prove the following:
Let $n\in \mathbb{N}$. Prove that $\not \exists$ a holomorphic function $f$ on the open unit disk satisfying:
$f\left(\displaystyle \frac{1}{n}\right) = 2^{-n}$
Any help will be deeply appreciated!
 A: Any function $f$ analytic in the unit disk and not identically $0$ can be written in the form
$$f(z)=z^r\ g(z)\ ,$$
where $r\in{\mathbb N}_{\geq0}$, $\ g$ is analytic in the unit disk, and $g(0)=:c\ne0$. Assume that for such an $f$ we have
$$2^{-n}=f\left({1\over n}\right)=n^{-r}\ g\left({1\over n}\right)\qquad (n\geq1)\ ,$$
or $g\bigl({1\over n}\bigr)=\ n^r\  2^{-n}$. Letting $n\to \infty$ we obtain
$$c=\lim_{n\to\infty} g\left({1\over n}\right)=\lim_{n\to\infty}\bigl(n^r\ 2^{-n}\bigr)=0\ ,$$
a contradiction.
A: Hint: Assume that $f$ is analytic on the open unit disk, then think about power series.  Show that $f(z)/z^k \to 0$ as $z \to 0$ for all fixed integers $k \geq 0$.
A: $\bf{Hint}$: Suppose we had such an analytic function $f$ on the open unit disk. Then in order for $f$ to even be continuous, $f(0) = 0$. Since $f$ is analytic we can express $f$ as a power series centered around zero $$f(z) = \sum_{n = 0}^{\infty}{\frac{f^{n}(0)}{n!}z^n}$$ Show that $f^n(0) = 0$ for all $n\in\mathbb{N}$. Deduce that $f$ cannot be analytic
