Does inverse of a nontrivial holomorphic function always have a branch point? Any nontrivial (i.e. which is not a first order polynomial) entire in $\mathbb{C}$ function I have thought of has a multifunction as its inverse and has a branch point. For example, $x^n\to\sqrt[n]{x}$, $\exp(x)\to\ln x$, etc.
So I have a question: are there any nontrivial entire functions, inverses of which have no branch points?
EDIT: the inverse may have other types of singularities - it needn't be holomorphic.
 A: Indeed, the only entire one-to-one functions are affine maps of the form $az+b$ with $a \neq 0$. To see this, suppose that $f:\mathbb{C} \to \mathbb{C}$ is a one-to-one entire function, and consider $g(z):=f(1/z)$. I claim that $0$ is a pole of $g$. Indeed, $0$ cannot be a removable singularity of $g$, for otherwise $g$ would be bounded near $0$, i.e. $f$ would be bounded near $\infty$, and thus $f$ would be constant by Liouville's Theorem. Also, $0$ cannot be an essential singularity of $g$. Indeed, if this was the case, then by the Casorati–Weierstrass theorem, $g(\mathbb{D} \setminus \{0\})$ would be dense in $\mathbb{C}$. But since $g$ is one-to-one, $g(\mathbb{C} \setminus \overline{\mathbb{D}})$ is an open set disjoint from $g(\mathbb{D} \setminus \{0\})$, a contradiction.
Therefore, $g$ has a pole at $0$. Furthermore, in order for $f$ to be entire, $g$ must be of the form
$$g(z)=a_0 + \frac{a_1}{z}+ \dots + \frac{a_n}{z^n}.$$
Hence $f$ is a polynomial, and since $f$ is one-to-one on $\mathbb{C}$, $f$ has to be of the form
$$f(z)=a_0z+a_1$$
for $a_0,a_1 \in \mathbb{C}$ and $a_0\neq0$.
A: There are no nontrivial entire functions whose inverse is holomorphic. More precisely,

If $D$ is an open, simply connected subset of $\mathbb C$ and  $f:\mathbb C \to D$ is biholomorphic, then $D=\mathbb C$ and $f$ is a polynomial of degree 1.

This is a consequence of the Riemann mapping theorem.
See for instance WHAT IS...a Biholomorphic Mapping?.
