$f(z) = w$ has n solutions $\forall w \in \mathbb{C} \iff f$ is a polynomial of degree $n$ Here is a conjecture that I came up with yesterday.
Conjecture: Suppose that $f:\mathbb{C} \to \mathbb{C}$ is entire. Then $f(z) = w$ has $n$ solutions $\forall w \in \mathbb{C} \iff f $ is a polynomial of degree $n$.
The converse follows immediately from the Fundamental Theorem of Algebra, but the forward implication is a lot more difficult. And after thinking about it more,  it seems to me that the forward implication is probably not true. Unfortunately, I can't immediately think of a counterexample.
I thought a little about the approach I might take to prove this. Suppose $f$ is as above. Then $f(z)=0$ has $n$ solutions, so we can write $f(z)=p(z)g(z)$ where $p$ is a polynomial of degree $n$ and $g$ is entire with $g\neq 0$. If $g$ is a constant, then we are done. Otherwise, by Liouville and Casaroti-Weierstrass, $g$ is unbounded and gets arbitrarily close to any value in $\mathbb{C}$. We can also write $g=e^h$ for some $h$ entire. (There is also Picard's theorem, but I am reluctant to use this as I don't know the proof.)
The above gets somewhere, but I am not sure it leads to a proof without another insight. If the conjecture is false, then it definitely doesn't lead to a proof.
So, is the conjecture true or false? And if it is true, am I on the right track to prove it? (I would appreciate small hints much more than a full answer.)
 A: One can easily prove this without Picard as the assumptions imply that (if $f$ nonconstant so $n \ge 1$) then $f$ is a proper entire map since for any compact $K$ then $f^{-1}(K)$ is bounded
(pick $w \in K$ there is a small open neighborhood of it $B_w$ for which $f^{-1}(B_w)$ is a union of at most $n$ open bounded sets - there are exactly $n$ precisely when $w$ is not a critical value of $f$ - cover $K$ with finitely many such $B_w$ etc)
But any proper entire map is a polynomial since then $|f(z)| \to \infty$ as $z \to \infty$
(if there is $|z_n| \to \infty$ st $|f(z_n)|$ bounded, one can pick a subsequence and rename it $z_n$ also, for which $f(z_n) \to w$ and then for a small ball $B_w$ around $w$, one has $f^{-1}(\overline B_w)$ containing elements with unbounded absolute value, hence it is not compact)
Edited: as per comments more detail about why $f$ is proper:
If $w$ is not a critical value of $f$, then by the local form of holomorphic functions, for every $z_k, k=1,..n$ in its preimage, there is a small neighborhood $D_k$ of $z_k$ for which $f$ is injective and has image some open neighborhood $U_k$ of $w$ so one can take $B_w$ any ball in the intersection of the $U_k$ which is open and contains $w$ - and then the preimage of $B_w$ is included in the union of $D_k$ as each contains a root of $f(z)=y$ for every $y \in B_w$, so we account for all $n$ of them; if $w$ is a critical value then for every preimage for which $f'(z_k)=0$ the same applies except that now $f$ is a $m:1$ map from $D_k$ onto $U_k$ where $m$ is the order of the critical point $z_k$ so in other words the order of the zero for $f(z)-f(z_k)$
$K$ is compact and covered by the $B_w$ so by the property of compacity, it is covered by finitely many such.
