Let $a_n$ be a sequence of complex numbers without an accumulation point. Let $c_n$ be an arbitrary sequence of complex numbers. Prove that, there is an entire function f, such that $f(a_n)=c_n$.
So the Weierstrass factorization ensures the existence of a holomorphic function, such that $f(a_n)=0$, so $c_n=0$, $\forall n$, we would be finished. But how can I proceed when $c_n \neq 0 \forall n$?
Wlog assume that $|a_n|$ is increasing. The $a_n$ are distinct. Set $f_1(z)=c_1$ and
$$f_k(z)= f_{k-1}(z)+ (c_k-f_{k-1}(a_k))(z/a_k)^{e_k} \prod_{n=1}^{k-1}\frac{z-a_n}{a_k-a_n}$$
To ensure that $\lim_{k\to \infty}f_k$ converges locally uniformly, take $e_k$ large enough so that $$\sup_{|z|<|a_k|-1}|(c_k-f_{k-1}(a_k))(z/a_k)^{e_k} \prod_{n=1}^{k-1}\frac{z-a_n}{a_k-a_n}|<2^{-k}$$