holomorophic function on unit disk

Question

Let $f$ holomorophic function on $\{|z|<1\}$. Suppose $|f(\frac{1}{n})|\leq e^{-n}$ for $n=2,3,\dots $ ,then $f=0$.

If $f(\frac{1}{n})=0$, by identity theorem, $f=0$.But I cannot prove it.

Answer 1

Under this hypothesis, then $z\mapsto e^{\frac1z}f(z)$ has a removable singularity at $0$. Therefore, if $f\neq0$, $e^{\frac1z}$ has a removable singularity or a pole at $0$. However, this is not true: $e^{\frac1z}$ has an essential singularity there.