If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$ for $|z|\geq R$, then $f$ is a polynomial of degree atleast $n$. I have a question in my assignment :

If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$
  for $|z|\geq R$ for some $n\in\mathbb N$ and some $M$ and $R$ in
  $(0,\infty)$ show that $f$ is a polynomial of degree atleast $n$.

Now , I defined a function    $\ g(z) = \frac {1}{f(z)}\ $ such that $\ |g(z)| \le \frac{1}{M{|z|}^n}.$
Now , by using the cauchy inequality 
$$|g^{(n)}(z)| \le  \frac{n!}{R^n |z|^nM}.$$ 
Considering that $ g(z) $ is an analytic function , it has a radius of convergence $ \infty $
$ \implies\  g^{(n)}(z) = 0.$ 
But if we go by this approach , then $ g^{(n)}(z) = 0 \ $ for any n . Also how can we be so sure that $ f(z) \neq 0 $ for any z ?
Is my reasoning correct or is there some other way to solve it ?
 A: As @Brian points out, $f$ has only finitely many zeros. Of course, $f(z)\neq 0$ if $|z|\geq R$. Since the set $B_R=\{z\mid |z|\leq R\}$ is compact, $f$ can only have finitely many zeros in $B_R$(use the identity theorem). Let $a_1,\ldots,a_k$ be the zeros of $f$ counted according to multiplicty. Let $$p(z)=(z-a_1)\cdots(z-a_k)=z^k+b_{k-1}z^{k-1}+\cdots+b_0.$$ For $|z|\geq R,$ we have $$|p(z)|\leq |z|^k\Bigl(1+\frac{|b_{k-1}|}{|z|}+\cdots+\frac{|b_{0}|}{|z|^k}\Bigl)\leq C|z|^k,$$ where $C=1+\frac{|b_{k-1}|}{R}+\cdots+\frac{|b_{0}|}{R^k}.$ Thus we have $$\frac{|z|^n|p(z)|}{|f(z)|}\leq \frac{|p(z)|}{M}\leq \frac{C|z|^k}{M},$$ for $|z|\geq R$. 
Suppose that $n=k$. Then, by Liouville, we see that $\frac{p(z)}{f(z)}$ is a constant function and hence $f$ is a polynomial of degree $k=n$. 
Suppose now that $n\lt k$. Then it is easy to see that $\frac{p(z)}{f(z)}$ is a polynomial of degree $\leq k-n$ (use the Cauchy's integral formula for derivatives. Click here for a proof.)
But $\frac{p(z)}{f(z)}$ is a nowhere vanishing entire function. So $\frac{p(z)}{f(z)}$ is a constant and hence $f$ is a polynomial of degree $k\gt n$.
Finally, assume $n\gt k$. Then, by Liouville's theorem, $\frac{z^{n-k}p(z)}{f(z)}$ is a constant. So $f(z)=cz^{n-k}p(z)$ for some constant $c$ and degree of $f$ is $n$. But $f$ and $p$ share the same zeros with same multiplicities. So degree of $f$ is equal to degree of $p$, i.e., $n=k$, a contradiction. (One can also use the Rouche's theorem to conclude. See @N. S.'s comment below.) 
A: Let $g(z) = f(1/z), z\in \mathbb C \setminus \{0\}.$ Then $|g(z)|\ge M/|z|^n$ for $0<|z|<1/R.$ Now $g$ blows up at $0,$ so has a nonremovable singularity there. Thus $g$ has an essential singularity at $0,$ or a pole at $0.$ If the former, then Casorati-Weierstrass tells us $g(\{0<|z|<1/R\})$ is dense in $\mathbb C.$ But $ g(\{0<|z|<1/R\})\subset \{|z|>MR^n\},$ contradiction.
Therefore $g$ has a pole at $0.$ Thus there is a polynomial $p$ and an entire $h$ such that $g(z) = p(1/z)+h(z)$ in $\mathbb C\setminus \{0\}.$ Flipping back, we see $f(z)=p(z) + h(1/z)$ on $\mathbb C\setminus \{0\}.$ It follows that as $|z|\to \infty,$ $f(z)-p(z) \to h(0).$ Because an entire function with finite limit $L$ at $\infty$ equals $L$ everywhere (Liouville), we arrive at $f(z) - p(z) = h(0)$ everywhere. The growth rate of $f$ then implies $p$ has degree at least $n$ as desired.
A: Definitions
Consider the $\{z_k\}$ where $f(z_k)=0$. Since they all must be in $|z|\le R$, if there were infinitely many, there would be a limit point and then, by the Identity Theorem, $f$ would be identically $0$.  At each $z_k$, there is an $d_k\in\mathbb{N}$, so that $f(z)=(z-z_k)^{d_k}g_k(z)$, where $g_k(z_k)\ne0$ and $g_k$ is entire. Therefore,
$$
g(z)=\frac{f(z)}{\prod\limits_{k=1}^m(z-z_k)^{d_k}}\tag2
$$
is entire yet does not vanish. Since $|g(z)|\gt0$, we must have $|g(z)|\ge L$ on $|z|\le R$ (since $|g|$ is a continuous function and $|z|\le R$ is a compact set, $|g|$ attains its infimum on $|z|\le R$).
On $|z|\gt R$,
$$
\begin{align}
\prod_{k=1}^m|z-z_k|^{d_k}
&\le\prod_{k=1}^m(|z|+|z_k|)^{d_k}\\
&\le\left[\prod_{k=1}^m\left(1+\frac{|z_k|}R\right)^{d_k}\right]|z|^d\\[6pt]
&=C|z|^d\tag3
\end{align}
$$
where $d=\sum\limits_{k=1}^md_k$.
Note that since $|z_k|\le R$, we have $C\le2^d$. 

Show that $\boldsymbol{d\ge n}$
Inequalities $(1)$ and $(3)$ say that
$$
|g(z)|\ge\frac MC|z|^{n-d}\tag4
$$
for $|z|\gt R$.
Let $h(z)=\frac1{g(z)}$, then
$$
|h(z)|\le\left\{\begin{array}{}
\frac1L&\text{for }|z|\le R\\
\frac CM|z|^{d-n}&\text{for }|z|\gt R
\end{array}\right.\tag5
$$
Suppose $d\lt n$, then $h(z)$ is bounded and entire. Thus, by Liouville's Theorem, $h$, and therefore $g$, would be constant. This implies that
$$
\begin{align}
\frac{|f(z)|}{|z|^n}
&=\frac{|g(0)|}{|z|^{n-d}}\prod_{k=1}^m\left|\frac{z-z_k}z\right|^{d_k}\\
&\hspace{-6pt}\overset{|z|\to\infty}\to0\tag6
\end{align}
$$
which contradicts $(1)$. Therefore, $d\ge n$.

Show that $\boldsymbol{h}$ and $\boldsymbol{g}$ are Constant
For $|z|\gt R$, $(5)$ says that $|h(z)|\le\frac CM|z|^{d-n}$. Thus, for $r\gt R$, Cauchy's Integral Formula says
$$
\begin{align}
\left|h^{(k)}(0)\right|
&=\frac{k!}{2\pi}\left|\int_{|z|=r}\frac{h(z)}{z^{k+1}}\mathrm{d}z\,\right|\\
&\le\frac{Ck!}Mr^{d-n-k}\tag7
\end{align}
$$
So if $k\gt d-n$, we have $h^{(k)}(0)=0$. That is, $h$ is a polynomial of degree at most $d-n$. However, if $h$ has degree greater than $0$, it would have a root, which would be a pole for $g(z)$, and therefore, $g$ would not be entire. So $h$ and $g$ must be constant.

Conclusion
Since $g$ is a constant,
$$
f(z)=g(0)\prod\limits_{k=1}^m(z-z_k)^{d_k}\tag8
$$
Therefore, $f$ is a polynomial of degree $d\ge n$.
