An entire function with pole at infinity assumes every value at only finitely many points in $\Bbb C$ 
Let  $f:\Bbb C\to \Bbb C$ be an entire function and $f$ has a pole at infinity. Show that $f$ assumes every value at least once, and at most finitely many times.

By Identity theorem $f$ will assume values only finitely many points(If possible give some other proof) but I am unable to see why it will assume every value. I am looking for a direct proof without using some heavy machinery like "Picard theorem" etc.Please help !
 A: $f$ has a pole at infinity means $\lim_{z->\infty} f(z) = \infty$ ...suppose $f(z)=x$ for infinitely many $z$ then there exists a compact set $B\subset \mathbb{C}$ s.t $f(z)\neq x$ for $z\in B^c$...now since $B$ is compact so there exists a limit point...so by identity theorem $f$ is constant so it has no pole in infinity.
suppose there exists $x\in \mathbb{C}$ s.t $f(z)\neq x$ for any z in $\mathbb{C}$...then claim there exissts an open set $N$ around $x$ in $\mathbb{C}$ s.t $f(\mathbb{C})\cap N=\phi$...if not then there exists a sequence ${z_n}$ in $\mathbb{C}$ s.t  $\lim_{n->\infty} f(z_n) =x$ now since $f$ has a pole in infinity so the seq ${z_n}$ is bounded so it has a  convergence subsequence  ..and since $f$ is continuous so by rule of continuity there exists some $z$ s.t $f(z)=x$...contradiction
now consider $F(z)=1/(f(z)-x)$ then F is bounded holomorphic entire function so constant...so $f$ is constant so it has no pole at infinity...contradiction 
A: I'm not sure if this is a valid proof, but let's give it a try: 
Assume value $a$ is taken infinitely many times. If $f$ has a pole at infinity, then outside of some closed disc around the origin the values of the function must have strictly larger absolute value than $a$ has. Thus the set $A=\{z\in\Bbb C:f(z)=a\}$ is infinite (by assumption) and bounded, so by Bolzano-Weierstrass $A$ has an accumulation point. But then, by identity theorem, because on $A$ this function takes the same value as constant function $g(z)=a$, the two functions must be equal, which is a contradiction, as $g$ doesn't have a pole at infinity.
A: Entire functions with poles at infinity are exactly the non constant polynomials.  Thus the result follows from the fundamental theorem of algebra.  (see Singularity at infinity of a function entire)
