TITLE: $g\circ f$ non-constant polynomials then $f$ and $g$ are polynomials?
QUESTION [8 upvotes]: Let $f$, $g$ be non-constant entire functions. If the composition  $g\circ f$ is a non-constant polynomial, can we conclude $f$ and $g$ are polynomials?

REPLY [3 votes]: Some basic properties: 1. If $h$ is entire and nonconstant, then $h(\mathbb C)$ is dense in $\mathbb C.$ (Casorati-Weierstrass and the fundamental theorem of algebra). Thus if $D$ is dense in $\mathbb C,$ then $h(D)$ is dense in $\mathbb C.$

If $p$ is a nonconstant polynomial, then for every $R>0,$ there exists $R'>0$ such that $p(\{|z|>R\})$ contains $\{|z|>R'\}.$ (A proof of this is sketched in the comments.)

Now in our problem $g\circ f$ is a nonconstant polynomial. Thus $(g\circ f) (\{|z|>R\})$ is not dense in $\mathbb C$ for some $R>0.$
If $f$ is not a polynomial, then $f (\{|z|>R\})$ is dense in $\mathbb C$ ((Casorati-Weierstrass). Hence $g(f (\{|z|>R\}))$ is dense in $\mathbb C$ by 1. above, conradiction. Thus $f$ is a polynomial.
By 2., this implies $f (\{|z|>R\})$ contains some $\{|z|>R'\}.$ We can assume $R'>R.$ If $g$ is not a polynomial, then it maps $\{|z|>R'\}$ to a dense subset of $\mathbb C$ (Casorati-Weierstrass). Hence $g\circ f$ maps $\{|z|>R\}$ to a dense subset of $\mathbb C.$ This is a contradiction, proving $g$ is a polynomial.