[proofplan] Part (1) uses Cauchy's estimate on $|z| = R$ to bound the Taylor coefficients $|a_n| \leq M(1+R)^k/R^n$, then sends $R \to \infty$ to kill all coefficients with $n > k$. Part (2) uses the Casorati-Weierstrass theorem: if $f$ were entire and transcendental with $|f(z)| \to \infty$, then $g(z) = f(1/z)$ would have an essential singularity at $0$ with dense image near $0$, contradicting $|g(z)| \to \infty$. [/proofplan] [step:Bound the Taylor coefficients using Cauchy's estimate to prove Part (1)] Write $f(z) = \sum_{n=0}^{\infty} a_n z^n$, which converges on all of $\mathbb{C}$ since $f$ is entire. By [Cauchy's Estimate](/theorems/346) applied on the circle $|z| = R$: \begin{align*} |a_n| \leq \frac{\sup_{|z| = R}|f(z)|}{R^n} \leq \frac{M(1 + R)^k}{R^n}. \end{align*} For $n > k$, the exponent of $R$ in the denominator exceeds that of the numerator: \begin{align*} \frac{(1 + R)^k}{R^n} \leq \frac{(2R)^k}{R^n} = \frac{2^k}{R^{n-k}} \to 0 \quad \text{as } R \to \infty. \end{align*} Hence $a_n = 0$ for all $n > k$, so $f$ is a polynomial of degree at most $k$. Conversely, if $f(z) = \sum_{n=0}^{k} a_n z^n$ is a polynomial of degree at most $k$, then by the triangle inequality: \begin{align*} |f(z)| \leq \sum_{n=0}^{k} |a_n||z|^n \leq \left(\sum_{n=0}^{k}|a_n|\right)(1 + |z|)^k. \end{align*} Setting $M = \sum_{n=0}^{k}|a_n|$ gives the growth bound. [/step] [step:Use Casorati-Weierstrass to prove Part (2)] If $f$ is a non-constant polynomial of degree $d \geq 1$, then $|f(z)| \geq \frac{|a_d|}{2}|z|^d$ for $|z|$ sufficiently large (by the dominance of the leading term), so $|f(z)| \to \infty$. Conversely, suppose $f$ is entire with $|f(z)| \to \infty$ as $|z| \to \infty$. Then $f$ is non-constant (a constant function cannot diverge). Assume for contradiction that $f$ is transcendental (not a polynomial). Define $g: \mathbb{C} \setminus \{0\} \to \mathbb{C}$ by $g(z) = f(1/z)$. Since $f$ is entire and transcendental, its Taylor series has infinitely many non-zero terms, so $g$ has an essential singularity at $z = 0$. By the [Casorati-Weierstrass Theorem](/theorems/???), for every $\varepsilon > 0$, the image $g(B(0, \varepsilon) \setminus \{0\})$ is dense in $\mathbb{C}$. In particular, there exists a sequence $z_m \to 0$ with $z_m \neq 0$ and $|g(z_m)| \leq 1$ for all $m$. But $|g(z_m)| = |f(1/z_m)|$, and $|1/z_m| \to \infty$, so $|f(1/z_m)| \to \infty$ by hypothesis. This contradicts $|g(z_m)| \leq 1$. Hence $f$ must be a polynomial. [guided] The reverse direction of Part (2) is the interesting one. We are given an entire function $f$ with $|f(z)| \to \infty$ as $|z| \to \infty$ and must show $f$ is a polynomial. The proof is by contradiction: if $f$ were transcendental, we would derive a contradiction using the behaviour of essential singularities. Define $g(z) = f(1/z)$ for $z \neq 0$. If $f$ is transcendental, its power series $\sum a_n z^n$ has $a_n \neq 0$ for infinitely many $n$, which means $g$ has an essential singularity at $z = 0$ (the Laurent series of $g$ about $0$ has infinitely many terms with negative powers). The [Casorati-Weierstrass Theorem](/theorems/???) states: if $h$ has an essential singularity at $z_0$, then for every $\varepsilon > 0$, $h(B(z_0, \varepsilon) \setminus \{z_0\})$ is dense in $\mathbb{C}$. Applying this to $g$ at $z_0 = 0$: for any $\varepsilon > 0$, the image $g(B(0, \varepsilon) \setminus \{0\})$ is dense in $\mathbb{C}$. In particular, for every $m \geq 1$, we can find $z_m \in B(0, 1/m) \setminus \{0\}$ with $|g(z_m) - 0| < 1$, i.e., $|g(z_m)| < 1$. But $z_m \to 0$ implies $|1/z_m| \to \infty$, so $|g(z_m)| = |f(1/z_m)| \to \infty$ by the growth hypothesis on $f$. This contradicts $|g(z_m)| < 1$ for all $m$. The contradiction forces $f$ to be a polynomial. [/guided] [/step]