The proof of the day is: $\mathbb{C}$ is algebraically closed
Theorem: The field $\mathbb{C}$ is algebraically closed.
Lemma 1: Every bounded everywhere-holomorphic function $f:\mathbb{C}\to\mathbb{C}$ is constant.
By assumption, there is $r \geq 0$ such that $|f(\mathbb{C})| \subset B_r$, where $B_r$ is the closed ball of radius $r$. For any positive real number $R$, consider $f_R(z) = f(Rz)$. By the Cauchy Integral Formula:
$$
|f'_R(z)| = \frac{1}{2\pi}\left|\int_{C_1(z)}\frac{f_R(w)}{(w-z)}dw\right|\leq\frac{1}{2\pi}\int_{C_1(z)}r|dw|=r
$$
Where $C_1$ is the circle of radius $1$ around $z$. Then:
$$
|f'(z)| = \frac{|f'_R(z)|}{R} \leq \frac r R
$$
Since $R$ was any real number, it follows that $|f'(\mathbb{C})|=\{0\}$, so $f$ is constant.
Proof:
Let $P$ be a $n$-degree polynomial over $\mathbb{C}$.
Suppose $P$ does not have zeroes in $\mathbb{C}$. So $\frac{1}{P(z)}$ is holomorphic in all of $\mathbb{C}$.
Additionally, there is always a positive real number $R$ such that:
$$
\left|\frac{1}{P(z)}\right| < \frac{2}{|a_n|R^n}\text{, Whenever $|z| > R$.}
$$
Where $a_n$ is the leading coefficient of $P$.
So, $\frac{1}{P(z)}$ is bounded in the region outside the disk $|z| \leq R$. However, $\frac{1}{P(z)}$is continuous on that closed disk, so it's bounded there as well, therefore $\frac{1}{P(z)}$ is bounded in $\mathbb{C}$.
By the Lemma 1, $P$ is constant. Contradiction.