Entire one-to-one functions are linear

Can we prove that every entire one-to-one function is linear?

You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.

To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f(\{z:|z|>n\})$ is dense in $\mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=\bigcap_n f(\{z:|z|>n\})$ is dense in $\mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.

I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f(\{z:|z|<1\})$ is open, and $f(\{z:|z|>1\})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.

By shifting $z$, without loss of generality you can assume $f(0) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 \over f(z)}$ is bounded outside of $U$. Therefore ${z \over f(z)}$ is an entire function that grows no faster than linearly: $|{z \over f(z)}| < A|z| + B$ for some $A$ and $B$.

It's easy from here to show that $g(z) = {z \over f(z)}$ is linear; for any $z_0$ ${g(z) - g(z_0) \over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z \over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {z\over c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 \over c_2} z$.

Let $f:\mathbb C\to\mathbb C$ entire and injective. Let $U=f(\mathbb C)$. $U$ is an open subset of the plane.

$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".

Next, if $U\subsetneq\mathbb C$, from Riemann's theorem we know that there is an biholomorphic map $U\to D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $\mathbb C\to D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.

Using this, one can see that the function $1/f(z)$ is bounded at $\infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.

(This avoids Picard but uses Riemann... :( )