how to show image of a non constant entire function is dense in $\mathbb{C}$?

It can be done with the help of the $\color{blue}{Casorati-Weierstrass\ theorem}$ or by the $\color{blue} {Liouville's\ theorem, }$

With the help of $\color{blue}{Casorati-Weierstrass\ theorem}$

Suppose $f$ is an entire function whose image is not dense in $\mathbb{C}$. Then there exists a complex number $\alpha$ and a number $s> 0$ such that |$f(z)-\alpha|>s$ $\forall z \in \mathbb{C}$.

Write $f(z) =\sum_{n=0}^{\infty} a_nz^n$ and suppose that there are infinitely many nonzero terms in this expansion. Then, for all $z \not=0$, let $g(z)=f(1/z)$ .We see that g has an essential singularity at $0$ so by the Casorati-Weierstrass theorem for some $z\ near\ 0$ we have $|g(z)-\alpha|<s $ then$|f(1/z)-\alpha|<s$ This contradiction implies that the power series expansion of $f$ can have only finitely many terms. Then the fundamental theorem of algebra guarantees that $f\ is\ constant$. This contradicts the hypothesis.

With the help of $\color{blue}{Liouville's\ theorem, }$

Suppose there exists a complex number $\alpha$ and $s\in \mathbb{R^+}$ such that $|f(z) — \alpha|>s, \forall z \in \mathbb{C}$. Then the function $g(z) = 1/(f(z) — \alpha)$ is entire and bounded, so by Liouville's theorem, g is constant. Hence $f$ is constant, again contradicting the hypothesis.

If the image weren't dense, it would miss a small disk. Inversion in the boundary of that disk would give you a non-constant bounded entire function, contrary to Liouville's Theorem.

To give a slightly more concrete rephrasing of Andreas's answer, if $f(z)$ is holomorphic and its image excludes $B(\lambda, r)$ for some $\lambda \in \mathbb{C}$, then $g(z) = \frac{1}{f(z) - \lambda}$ is entire and $|g(z)| \le 1/r$, so $g$ is constant, and hence $f$ is.