On every simply connected domain, there exists a holomorphic function with no analytic continuation. I am working on a question that requires me to prove that on every simply connected open subset of $\mathbb{C}$, there exists a holomorphic function that cannot be extended to a holomorphic function on a larger connected open set.
I know an example of such a holomorphic function on the open unit disc: $$f:z\mapsto\sum_{n=1}^\infty z^{n!}.$$ I tried to combine this with the Riemann mapping theorem.
Let $\Omega$ be a simply connected open subset of $\mathbb{C}$ and one may assume that $\Omega$ is not the entire complex plane. By the Riemann mapping theorem, there exists a conformal equivalence $$\phi:\Omega\rightarrow D$$where $D$ is the open unit disc. Then my guess is that $f\circ\phi$ has no analytic continuation, but then I had to link the boundaries of $\Omega$ and of $D$ and I don't know what is going on between $\partial\Omega$ and $\partial D$.
Could anyone offer any idea? Many thanks!
 A: I don't think this has much to do with simple connectivity. Let $U\subset \mathbb {C}$ be any open set with non-empty boundary. Claim: There exists $f\in H(U)$ such that
$$\sup_{D(a,r)\cap U} |f|=\infty$$
for every $a \in \partial U$ and $r>0.$ This $f$ cannot be be extended analytically to any larger open set containing a point in $\partial U.$
Proof: Let $\{a_1,a_2, \dots \}$ be a countable dense subset of $\partial U.$ The idea is to define
$$f(z) = \sum_{n=1}^{\infty} \frac{c_n}{z-a_n}$$
for appropriate positive constants $c_n.$
To do this, we choose pairwise disjoint countably infinite sets $E_1, E_2,\dots \subset U$ such that for each $n,$ the only limit point of $E_n$ in $\mathbb {C}$ is $a_n.$ (For each $n,$ $E_n$ is just a sequence of distinct points in $U$ whose limit is $a_n.$  You can choose the $E_n$'s inductively.)
Now there are compact sets $K_1,K_2, \dots \subset U$ such that
$$K_1\subset \text {int}(K_2)\subset K_2 \subset \text {int}(K_3) \subset \dots$$
such that $U= \cup K_n.$ Choose $c_1$ such that $c_1/|z-a_1| < 1/2$ on $K_1.$ If $c_1,\dots c_n$ have been chosen, we choose $c_{n+1}$ such that 
$$\frac{c_{n+1}}{|z-a_{n+1}|} < \frac{1}{2^{n+1}},\ \ z\in K_{n+1}\cup E_1 \cup \cdots E_n.$$
Then the series $\sum c_n/(z-a_n)$ converges uniformly on each $K_n,$ hence defines a holomorphic function $f$ on $U.$ Let $a\in \partial U, r > 0.$ Then $D(a,r)$ contains some $a_n,$ hence contains the tail-end of $E_n.$ Thus as $z\to a_n$ within $E_n,$ we have
$$|f(z)| \ge \left|\sum_{k=1}^{n}\frac{c_n}{z-a_k}\right| - \sum_{k=n+1}^{\infty}\frac{1}{2^{k+1}}\to \infty.$$
This completes the proof.
A: HINT:
The function $\sum_{n\ge 0}z ^{n!}$ can be seen to be un-extendable for an intrinsic reason: there exists a sequence $(a_n)$ of points in the disk such that $|f(a_n)| \to \infty$,  and every open subset of the disk that is not relatively compact contains a point $a_n$ ( so in fact infinitely many). 
Try to find such a sequence $(a_n)$. 
$\bf{Added}$ If $0 < z < 1$ and $z = r e^{2\pi \frac{k}{N}}$ then for all $n \ge N$ we have $z^{n!} = |z|^{n!}$. If you fix the argument $2 \pi\frac{k}{N}$ but increase the absolute value towards $1$ you see that $|f(z)| \to \infty$. Just choose $N$ points with arguments $0$, $2\pi \frac{1}{N}$, $\ldots $ , $2 \pi \frac{N-1}{N}$. such that say $f(\cdot)$ is in absolute value $> N$. 
A: Another proof of the claim of zhw is the following:
Given an open set $A$ and a set $B\subset A$ without cluster points in $A$, a slight generalization of Weierstrass factorization theorem (see here, or go to Rudin's "Real and Complex analysis" for a proof) ensure the existence of a function $f\in \mathcal{H}(A):\{z\in A|f(z)=0\}=B$.
Thanks to this theorem, we can easily prove the claim:
Choose as $B$ a sequence that accumulates to every element of $\partial A$: the associated $f$ cannot be analytically extended to a larger open set, as the identity principle would imply $f\equiv 0$.
