If $f$ is entire and $f(z)\to \infty,$ for $z\to \infty$ , then $f(z)$ is polynomial.
As $f(z)\to \infty, for z\to \infty$ then $\exists M,$ such that $|z|>M$ implies $|f(z)|>1$
COnsider COmpact set $\overline {B(0,M)}$ Then $f(z)$ has finite zeros in that compact set say $x_1,...,x_n$
Now $\frac{f(z)}{(z-x_1)...(z-x_n)}=g(z)$ is never vanishing function and tend to infinity.
ALso $h(z)=1/g(z)$ is holomorphic .
Now I can not abel to proceed further
Any Help will be appreciated
For $z \ne 0$ let $g(z):=f(1/z)$. Then $g$ hat an isolated singularity at $z=0$. From $g(z) \to \infty$ for $z \to 0$ we see that $g$ has a pole ate $z=0$.
Let $f(z)= \sum_{n=0}^{\infty}a_n z^n$, then $g(z)=\sum_{n=0}^{\infty}\frac{a_n}{z^n}$.
Since $g$ has a pole at $z=0$, there is $m \in \mathbb N$ such that $a_m=0$ for $m>n$.