Why in this case $f$ should be entire? Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a function.
Assume $f$ satisfies follows:


*

*$f$ is analytic at a point $z_0$.

*$\limsup\limits_{n\to\infty} \left|\frac{f^{(n)}(z_0)}{n!}\right|^{1/n}=0$.
Why in this case $f$ should be analytic everywhere?
That is, define $g(z)=\sum \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n$.
Why $g=f$ on $\mathbb{C}$?
What gurantees that they coincide even outside of a given neighborhood of $z_0$?
 A: As you note yourself, this is not true without additional assumptions on $f$. Take for example $z_0=0$,
$$g(z)=e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$$
and let $f$ be a function that is equal to $g$ on some (arbitrarily small) neighbourhood of $0$ and everywhere discontinuous elsewhere.
A: The fact that your function is analytic in $z_0$ means that there exists a radius $R>0$ s.t. 
$$
f(z)=\sum_{n=0}^{+\infty}c_n(z-z_0)^n
$$
holds for every $z\in B(z_0,R[$.
But... who is $R$? By definition $R$ is the largest possible in which $f$ is analytic; but in this way we don't say nothing useful.
The fact is that your second hypotesis guarantees the convergence of the power series on the ball $B(z_0,T[$ where $1/T=\limsup_n|c_n|^{1/n}=0$ so $T=+\infty$ and you have a power series that defines a function that is analytic on the whole complex plane (in fact $B(z_0,+\infty[=\mathbb C$). Call this function $g$. You now have two functions, $f$ and $g$, both analytic on a nhbrd on $z_0$ in which they coincide. Hence by the identity principle, they must coincide everywhere. But $g$ is defined on the whole complex plane, hence it will be THE (unique possible) analytic extension of $f$.
