[proofplan] We show that every Taylor coefficient of $f$ of order greater than $n$ vanishes, so that the Taylor series terminates and $f$ is a polynomial of degree at most $n$. The key tool is the [Cauchy Estimates](/theorems/2571): for a holomorphic function on a disc of radius $R$, the $k$-th Taylor coefficient is bounded by $M_R / R^k$, where $M_R$ is the supremum of $|f|$ on the boundary. The polynomial growth bound $|f(z)| \le C(1 + |z|^n)$ ensures that $M_R / R^k \to 0$ as $R \to \infty$ whenever $k > n$, forcing each such coefficient to be zero. [/proofplan] [step:Express $f$ as a convergent Taylor series centred at the origin] Since $f: \mathbb{C} \to \mathbb{C}$ is entire, by [Taylor's Theorem for Holomorphic Functions](/theorems/348), $f$ has a Taylor series expansion convergent on all of $\mathbb{C}$: \begin{align*} f(z) &= \sum_{k=0}^{\infty} a_k z^k, \qquad a_k = \frac{f^{(k)}(0)}{k!}, \quad z \in \mathbb{C}. \end{align*} It suffices to show that $a_k = 0$ for every $k > n$. [/step] [step:Apply the Cauchy Estimates to bound $|a_k|$ in terms of $R$] Fix an integer $k > n$ and let $R > 0$. Since $f$ is entire, $\overline{B}(0, R) \subset \mathbb{C}$, and the [Cauchy Estimates](/theorems/2571) give \begin{align*} |a_k| = \frac{|f^{(k)}(0)|}{k!} \le \frac{M_R}{R^k}, \end{align*} where $M_R = \sup_{|z| = R} |f(z)|$ is the supremum of $|f|$ on the circle $\partial B(0, R)$. [guided] The [Cauchy Estimates](/theorems/2571) require that $f$ be holomorphic on an open set containing $\overline{B}(0, R)$. Since $f$ is entire, it is holomorphic on all of $\mathbb{C}$, so this condition is satisfied for every $R > 0$. The estimate gives a bound on $|f^{(k)}(0)|$ in terms of the supremum of $|f|$ on the boundary circle: \begin{align*} |f^{(k)}(0)| \le \frac{k!\, M_R}{R^k}. \end{align*} Dividing both sides by $k!$ yields the bound on the Taylor coefficient: \begin{align*} |a_k| = \frac{|f^{(k)}(0)|}{k!} \le \frac{M_R}{R^k}. \end{align*} The strategy is now to exploit the growth hypothesis to show that $M_R / R^k \to 0$ as $R \to \infty$ whenever $k > n$. [/guided] [/step] [step:Use the polynomial growth bound to estimate $M_R$ and send $R \to \infty$] By hypothesis, $|f(z)| \le C(1 + |z|^n)$ for all $z \in \mathbb{C}$. In particular, for $|z| = R$: \begin{align*} M_R = \sup_{|z| = R} |f(z)| \le C(1 + R^n). \end{align*} Substituting into the Cauchy bound from the previous step: \begin{align*} |a_k| \le \frac{C(1 + R^n)}{R^k} = C\!\left(\frac{1}{R^k} + \frac{1}{R^{k-n}}\right). \end{align*} Since $k > n$, both exponents $k$ and $k - n$ are positive. Sending $R \to \infty$: \begin{align*} |a_k| \le \lim_{R \to \infty} C\!\left(\frac{1}{R^k} + \frac{1}{R^{k-n}}\right) = 0. \end{align*} Therefore $a_k = 0$ for every integer $k > n$. [guided] We use the growth hypothesis $|f(z)| \le C(1 + |z|^n)$ to control $M_R$. On the circle $|z| = R$, we have $|z|^n = R^n$, so \begin{align*} M_R = \sup_{|z| = R} |f(z)| \le C(1 + R^n). \end{align*} Substituting this into the Cauchy estimate $|a_k| \le M_R / R^k$ gives \begin{align*} |a_k| \le \frac{C(1 + R^n)}{R^k} = \frac{C}{R^k} + \frac{C}{R^{k-n}}. \end{align*} Now the key point: since $k > n$, the exponent $k - n > 0$, so $R^{-(k-n)} \to 0$ as $R \to \infty$. Similarly $R^{-k} \to 0$. The bound holds for every $R > 0$, so taking $R \to \infty$: \begin{align*} |a_k| \le \lim_{R \to \infty} \left(\frac{C}{R^k} + \frac{C}{R^{k-n}}\right) = 0. \end{align*} This forces $a_k = 0$. What would fail if $k \le n$? In that case $k - n \le 0$, and $R^{-(k-n)} = R^{n-k}$ would grow (or remain constant at $1$ if $k = n$) as $R \to \infty$, so the limit would not be zero. The argument is sharp: the growth bound $|f(z)| \le C(1 + |z|^n)$ exactly characterises polynomials of degree at most $n$. [/guided] [/step] [step:Conclude that $f$ is a polynomial of degree at most $n$] Since $a_k = 0$ for every $k > n$, the Taylor series of $f$ terminates: \begin{align*} f(z) &= \sum_{k=0}^{n} a_k z^k, \end{align*} which is a polynomial of degree at most $n$. This completes the proof. [/step]