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 $n$.

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.)