How many distinct holomorphic function from $\mathbb{C}$ to $\mathbb{C}$ is there?

Question

Is this proof OK ?

Considering that the constant function $f(z) = c$ is analytic for all $c \in \mathbb{C}$, there is at least $\mathfrak{c}$ (the power of the continuum) holomorphic function from $\mathbb{C}$ to $\mathbb{C}$.

Since a function from $\mathbb{C}$ to $\mathbb{C}$ is a subset of $\mathbb{C}\times \mathbb{C}$, there must be at most card$(\mathbb{C}\times \mathbb{C}) = \mathfrak{c}\cdot \mathfrak{c}$ functions from $\mathbb{C}$ to $\mathbb{C}$.

Assuming the axiom of choice, $\mathfrak{c}$ is an $\aleph$ and we know that the $\aleph$ satisfy $\aleph_\zeta \cdot \aleph_\zeta = \aleph_\zeta$. So there is at most $\mathfrak{c}$ holomorphic functions from $\mathbb{C}$ to $\mathbb{C}$.

Answer 1

Every entire function $\mathbb{C} \rightarrow \mathbb{C}$ is uniquely determined by its power series coefficients $\{a_0, a_1, \ldots, \}$.

Answer 2

Hint: A continuous function is determined by its values on a dense set.

Answer 3

Note that not every subset can be the range of an entire function, e.g., Liouville's theorem implies bounded sets must be singletons. That's why your argument doesn't work.