TITLE: Entire one-to-one functions are linear
QUESTION [65 upvotes]: Can we prove that every entire one-to-one function is linear?

REPLY [8 votes]: This proof uses only the Open Mapping Theorem and Cauchy's Estimates.
Suppose $f$ is entire and injective. Consider the injective entire function $$\widetilde{f} \colon z \mapsto f(z) - f(0).$$ We see that $\widetilde{f}(0) = 0$.  Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $\widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $\epsilon > 0$ such that $B(0,\epsilon) \subset \widetilde{f}(B(0,r))$. Since $\widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,\epsilon)$. In other words, $$|z| \geq r \Rightarrow |\widetilde{f}(z)| \geq \epsilon.$$
Now $\widetilde{f}$ has a zero at the origin, so let's say this zero has order $k \geq 1$. This means there is some entire function $g \colon \Bbb C \rightarrow \Bbb C$ such that $g(0) \neq 0$ and $$\widetilde{f}(z) = z^k g(z) \ \forall \ z \in \Bbb C.$$
Since $\widetilde{f}$ is injective, $\widetilde{f}$ has no other zeroes apart from its zero at the origin.  So $g$ has no zeroes.  This means the function $F \colon \Bbb C \rightarrow \Bbb C$ defined by $$F(z) = \frac{1}{g(z)} \ \forall \ z \in \Bbb C$$ is entire and also has no zeroes.  Write the Taylor expansion of $F$ at the origin as $$F(z) = \sum_{n=0}^\infty \frac{F^{(n)}(0)}{n!}z^n.$$
Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin.  Indeed,
$$F^{(n)}(0) \leq  \max_{|z| = R}|F(z)|\frac{n! }{R^n}.$$  But we can apply the fact that $|\widetilde{f}(z)| \geq \epsilon \ \forall \ |z| \geq r$ to notice that $$\max_{|z| = R} |F(z)| = \max_{|z| = R} \frac{1}{|g(z)|} = \max_{|z| = R} \frac{|z|^k}{|\widetilde{f}(z)|} \leq \frac{R^k}{\epsilon}.$$  So if $n > k$, we have $$F^{(n)}(0) \leq \max_{|z| = R}|F(z)|\frac{n!}{R^n} \leq \frac{R^k n!}{\epsilon R^n} = \frac{n!}{\epsilon R^{n-k}} \xrightarrow{R \to \infty} 0.$$
So we conclude from the Taylor expansion of $F$ that $$F(z) = \sum_{n=0}^k \frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes.  So $F \equiv c$ for some $c \in\Bbb C \backslash\{ 0\}$!  This means $g \equiv c^{-1}$, and furthermore, $$\widetilde{f}(z) = c^{-1}z^k \ \forall \ z \in \Bbb C.$$
Supposing $k \geq 2$, the polynomial $z^k - 1 \in \Bbb C[z]$ has at least two roots $\xi_1, \xi_2 \in \Bbb C$ which are certainly non-zero on account of the fact that $0^k \neq 1$. The multiplicity of the root $\xi_1$ is precisely $1$ because $$\frac{d}{dz}\Big\vert_{z=\xi_1} z^k - 1 = k \xi_1^{k-1} \neq 0.$$ This allows us to conclude that $\xi_1 \neq \xi_2$, and that $$\widetilde{f}(\xi_1) = c^{-1} \xi_1^k = c^{-1} \cdot 1 = c^{-1} \xi_2^k = \widetilde{f}(\xi_2).$$ This result contradicts the fact that $\widetilde{f}$ is injective.  So $k=1$, and our final conclusion is that $$f(z) = \widetilde{f}(z) + f(0) = c^{-1} z + f(0) \ \forall \ z \in \Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.