Let $f:\mathbb{C}\to\mathbb{C}$ be a bounded, entire function. Then by Taylor's theorem, $$f(z)=\sum_{n=0}^\infty c_n x^n \mbox{ where } c_n = \frac{1}{2\pi i}\int_{\Gamma_r} \frac{f(w)}{w^{n+1}}\,dw$$

where $\Gamma_r$ is the circle of radius $r$ about $0$ , for $r>0$ . Then $c_n$ can be estimated as $$|c_n|\leq\frac{1}{2\pi}\operatorname{length}(\Gamma_r)\operatorname{sup}\left\{\left|\frac{f(w)}{w^{n+1}}\right|\,\colon\, w\in \Gamma_r\right\} = \frac{1}{2\pi}\, 2\pi r \frac{M_r}{r^{n+1}} = \frac{M_r}{r^{n}}$$

where $M_r = \operatorname{sup}\{|f(w)|\colon w\in\Gamma_r\}$ .

But $f$ is bounded, so there is $M$ such that $M_r\leq M$ for all $r$ . Then $|c_n|\leq\frac{M}{r^n}$ for all $n$ and all $r>0$ . But since $r$ is arbitrary, this gives $c_n = 0$ whenever $n>0$ . So $f(z) = c_0$ for all $z$ , so $f$ is constant.$\square$