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 ?
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.)
Let $g(z) = f(1/z), z\in \mathbb C \setminus \{0\}.$ Then $|g(z)|\ge M/|z|^n$ for $0&lt;|z|&lt;1/R.$ 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 $g(\{0&lt;|z|&lt;1/R\})$ is dense in $\mathbb C.$ But $ g(\{0&lt;|z|&lt;1/R\})\subset \{|z|&gt;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.