Show that an entire function is a polynomial

There is a question in the book that asks me to show that if f is an entire function such that $|f(z)| \le L|z|^m$ where $|z| \ge R$, then $f$ is a polynomial of degree of at most $m$.

The problem gives me a hint that I should use the Cauchy estimates for n>m and $r \to \infty$

The below is from a post https://math.stackexchange.com/a/143881/64742

Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $\infty$, which must match its Taylor series there.

$$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$$

Since $|f(z)|\leq k|z|^m$, Cauchy's estimate gives us

$$|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}$$ for all $|z|=R$. For $n>m$, letting $R\rightarrow\infty$, we see that $|f^{(n)}|=0$. It follows that $f$ is a polynomial of degree $\leq m$.

Now, I follow what the above answer says except where "It follows that f is a polynomial of degree $ \le m$. Why do we arrive at that conclusion? The preceding statement merely says that $ |f^{(n)}|=0 $

Since $f$ is entire, for every $z\in\mathbb C$ it holds that $$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n \tag1$$ Once we know that $f^{(n)}(0)=0$ for $n>m$, the formula (1) simplifies to $$f(z)=\sum_{n=0}^m \frac{f^{(n)}(0)}{n!}z^n \tag2$$ where the expression on the right is evidently a polynomial of degree at most $m$.