Let $O\subset\mathbb{C}$ be open and $f:O\rightarrow \mathbb{C}$ an analytic function. Further let $a>0$ and $x\in O$ be such that $\overline{U_a(x)}$ is completely contained in $O$.
1 How do I show that $|\frac{d^k}{dz^k}f(x)|\leq\sup_{z\in\overline{U_a(x)}}|f(z)|\frac{k!}{a^k}$?
2 Let $g:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function such that $|g(z)|\leq A |z|^k$ for all $|z|>1$ and a constant $A$. How do I show that $g$ must be a polynomial in $z$ with degree $k$ or less?
What I know:
1 I was thinking maybe to use Cauchy generalized integral formula, but I wouldn't know how, or if I should at all.
2 My only idea here was to use the result from 1, but more than that I could not think of.
I would greatly apreciate any nudges in the right direction!
Hint For 1, differentiate under the integral in the Cauchy integral formula and then estimate using the M-L inequality. For 2, use 1 and the fact that $a>0$ can be as large as you like.
EDIT: By the Cauchy integral formula, it is not hard to show $$\frac{\mathrm d^k}{\mathrm d^kx}f(x)=\frac{k!}{2\pi i}\int_{|z-x|=a}\frac{f(z)}{(z-x)^{k+1}}\,\mathrm dz$$ Using the M-L inequality, we find $$\left|\frac{\mathrm d^k}{\mathrm d^kx}f(x)\right|\le\frac1{2\pi}\sup_{|z-x|=a}\left|\frac{f(z)}{(z-x)^{k+1}}\right|\cdot2\pi a\le ak!\cdot\sup_{z\in\overline{U_a(x)}}\frac{|f(z)|}{a^{k+1}}$$ which is what we want. Now for 2, replace $f$ by $g$ and $k$ by $k+1$. What happens when $a\to\infty$?