Find all entire functions $f$ such that $f\left( e^{-\frac{i\pi}{n}} \right) = e^{\frac{2\pi i}{n}} ,\ \forall n \in \mathbb{N}$
My proof:
Consider a sequence $a_n = \left\{ e^{ -\frac{i\pi}{n}} \right\}_{n \in \Bbb{N}}$ which has a limit in $\Bbb{C}$, namely $1$. And consider a function $g(z)=\frac1{z^2}$ which is holomorphic on $\Bbb{C} \setminus \{0\}$. For each point $a_n$ from the sequence the following holds: $$g(a_n) = \frac{1}{\left( e^{-\frac{i\pi}{n}} \right)^2} = e^{\frac{2i\pi}{n}} = f(a_n)$$ Now since $f$ and $g$ are holomorphic and equal at all points of a sequence which converges in $\Bbb{C} \setminus \{0\}$, and $\Bbb{C} \setminus \{0\}$ is an open and connected set, we can apply the identity theorem and conclude that $f(z) = g(z), \forall z \in \Bbb{C} \setminus \{0\}$.
But, alas, $g$ is not an entire function, so we can suppose now that there exists another function $h$ which is holomorphic on the entire $\Bbb{C}$ which satisfies our condition. So this function would have to be equal $f$, but that means it is also equal to $g$ on $\Bbb{C} \setminus \{0\}$, but there is no way to define $g$ at zero so that it is holomorphic since zero is a pole of second order for this function.
Hence we conclude that such function $h$ does not exist, which finally means that there cannot be a function $f$ which is both entire and satisfies the given condition.
Is this a valid reasoning?