I want to prove that if $f$ is an entire function and $f(z) \not \in [0,1]$ for every $z$, then $f$ is constant.
If it was written that $|f(z)| \not \in [0,1]$ I would have used the fact that $\frac{1}{f}$ is entire function and Cauchy's formula to bound it.
However it is not the case here, so I don't really see what to do.
I can still use the fact that $\frac{1}{f}$ is also entire, but appart from that, I am clueless.
Help would be appreciated
Without using Picard's theorem, the idea is to find an injective holomorphic function $\phi:\mathbb{C}\setminus[0,1]\to\mathbb{D}$, where $\mathbb{D}$ is the (open) unit disk. It then follows that $\phi\circ f$ is bounded and entire, hence constant, and so $f$ is constant since $\phi$ is injective.
To find $\phi$ it's probably easiest to use a composition of multiple functions, say first mapping $\mathbb{C}\setminus[0,1]$ into $\mathbb{C}\setminus[0,\infty)$, then into an open halfplane, then into $\mathbb{D}$.
Picard's little theorem: A non-constant entire function $f$ takes all values of complex plane with at most a single exception.
Further, if it misses more than one value then $f$ is constant.
$ Added$:Define $g:\mathbb C\rightarrow \mathbb C$ as $g=\frac{1}{f}$. Clearly, $g$ misses at least two values ( 1 and 2 say) on $\mathbb C$ (codomain plane),so $g$ must be constant by Picard's little theorem and so is $f$.
