 Solution

Use the boundary form of Cauchy's integral theorem: the integral of a holomorphic function around the oriented boundary of a compact region on whose neighborhood it is holomorphic is zero, including annular regions with the inner boundary oppositely oriented. Apply it to $f(z)/(z-a)$ between the circles of radii $r$ and $\rho$, both around $a$. It givesas $\rho \downarrow 0$, by continuity. This proves the Cauchy integral formula.

For a bounded entire function with $|f|\le M$, apply that formula at variable points inside a fixed circle and differentiate its kernel under the integral. At its center $a$, this givesThe circle can have any radius because the function is entire. Letting $R\to \infty$ gives $f^{\prime }(a)=0$ at every $a$, so $f$ is constant. This proves Liouville's theorem.

Finally, if a nonconstant entire function omitted all points within distance $ϵ$ of some $w$, then $g=1/(f-w)$ would be entire and bounded by $1/ϵ$. Liouville's theorem would make $g$, and hence $f$, constant. Therefore every such neighborhood contains a value of $f$, proving the dense image of a nonconstant entire function property.