Let $g:\mathbb{C} \to \mathbb{C}$ an entire function. Prove that the Taylor expansion around $0$ converges uniformly in all $\mathbb{C}$ if and only if $g$ is a polynomial.
1/2 PROOF I think I have successfully proved the "if " part since if $g$ is a polynomial, for some $m \in \mathbb{N}$ we have $$ g(z)=a_mz^m+a_{m-1}z^{m-1}+ \cdots + a_1z+a_0. \ \ $$ The taylor expansion takes the form $$ \sum_{n=0}^{\infty} b_n z^n $$ with $b_n=g^{(n)}(0)/n!$. Differentiation of $g$ gives that $b_n=a_n$ for $n=0,1,\cdots, m$ and $b_n=0$ if $n>m$. Then if $k\geq m$ and $z \in \mathbb{C}$ $$ \left| \sum_{n=0}^{k} b_n z^n - g(z) \right| = 0 $$ so , indeed, $\sum_{n=0}^{k} b_n z^n$ converges to $g$ uniformly.
PROBLEM Now, to prove the converse, I guess I have to do it by contrapositive and use the fact that $g$ is an entire function. However, I am pretty lost, and I think I am missing something very obvious since I am supposed to answer this with only the first 3 chapters of the book "Conway: Functions of One Complex Variable I ". Any help or Hints will be very appreciated.
Thanks to @BillTrok for the hint. Here is the answer:
$\boldsymbol{(\Longrightarrow)}$ If the Taylor expansion of $g$, around 0 converges uniformly in all $\mathbb{C}$ to $g$, put $T_k(z)=\sum_{n=0}^{k} b_n z^n$, then for all $\varepsilon>0$ there exist $N=N(\varepsilon) \in \mathbb{N}$ such that if $k\geq N$ and $z \in \mathbb{C}$ then $$ |T_k(z)-g(z)|<\varepsilon $$ So, if we take $k\geq N$ $$ |b_{k+1}z^{k+1}|=|T_{k+1}(z)-T_{k}(z)| \leq |T_{k+1}(z)-g(z)|+|T_k(z)-g(z)|< 2\varepsilon \ \ \forall \ z \in \mathbb{C} $$ this gives that $|b_{k+1}z^{k+1}|$ is arbitrary small for larges $|z|$, then we must have $b_{k+1}=0$ for all $k \geq N$, which gives that $g$ is in fact a polynomial with degree $N$.