Suppose that $f(z)$ is an entire function such that $f(z)/z^n$ is bounded for $|z|\geq R$. SHow that $f(z)$ is a polynomial of degree at most $n$.
As $f(z)$ is an entire function i thought of considering power series expansion $$f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n$$
But then this question is asked before introducing power series.
I checked if i can do by using this.
My aim is to prove that $f^{(d)}(0)=0$ for all $d\geq n+1$.
Estimate $$f^{(d)}(0)=\frac{n!}{2\pi i}\int_{C(0,R)}\frac{f(z)}{z^{d+1}}dz$$
Then i thought i can use that $f(z)/z^n$ is bounded. But then we have that for $z:|z|>R$
Help me in estimating
Hint: Let $R'>R$ and consider $$ \frac{(n+1)!}{2\pi i}\int_{C(0,R')}\frac{f(z)}{z^{n+2}}dz. $$ Since you are integrating on the circle of radius $R'$ and $R'>R$ this is not an integral over a disk, you know that $\left|\frac{f(z)}{z^{n+2}}\right|\leq\frac{C}{(R')^2}$ for some constant $C$.
Taking the magnitude of the integral, $$ \left|\frac{(n+1)!}{2\pi i}\int_{C(0,R')}\frac{f(z)}{z^{n+2}}\,dz\right|\leq\frac{(n+1)!}{2\pi}\int_{C(0,R')}\left|\frac{f(z)}{z^{n+2}}\right|\,|dz|\leq\frac{(n+1)!}{2\pi}\int_{C(0,R')}\frac{C}{(R')^2}\,|dz|. $$
Since the length of a circle of radius $R'$ is $2\pi R'$, you have that $$ \left|\frac{(n+1)!}{2\pi i}\int_{C(0,R')}\frac{f(z)}{z^{n+2}}dz\right|\leq\frac{(n+1)!}{2\pi}\int_{C(0,R')}\frac{C}{(R')^2}\,|dz|\leq\frac{(n+1)!}{2\pi}\cdot\frac{2\pi C\cdot R'}{(R')^2}. $$
Since $R'$ is arbitrary (but larger than $R$), this inequality holds for all $R'$ large enough. By letting $R'$ go to infinity, you can conclude that $f^{(n+1)}(0)=0$.
Now, generalize for other points and think about what it means that the $(n+1)$st derivative is identically $0$.