Let $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$. Let's consider the analytic function $f:\mathbb{D}\to\mathbb{C}$ given by, for all $z\in\mathbb{D}$, $$f(z)=\prod_{n=1}^\infty (1-z^n)^{-1}.$$ For each positive integer $n$, let $p(n)$ be the number of partitions of $n$ into positive integers. Then, for $z\in\mathbb{D}$, $f$ admits the power series expansion $$f(z)=1+\sum_{n=1}^\infty p(n)z^n.$$ By the monotone convergence theorem, $$\lim_{k\to\infty}f\left(\frac{k}{k+1}\right)=\lim_{k\to\infty}\sum_{n=1}^\infty p(n) \left(\frac{k}{k+1}\right)^n=\sum_{n=1}^\infty \lim_{k\to\infty} p(n) \left(\frac{k}{k+1}\right)^n=\sum_{n=1}^\infty p(n) =\infty.$$ Therefore, $f$ cannot be analytically continued to the point $1$. It is well-known (but rarely proved) that
the boundary of $\mathbb{D}$ is a natural boundary for $f$.
Actually, come to think of it, I haven't even found a proof of this fact. I would very much appreciate a reference or a rigorous proof.
In Remmert, Funktionentheorie 2 / Classical Topics in Complex Function Theory, Chapter 11 there is a
Theorem by Gábor Szegő. Let $f=\sum a_n z^n$ a power series such that there are only finite many different values for the coefficients. Then either $\mathbb{D}$ is the natural boundary for $f$ or $f(z)=\frac{p(z)}{z^N-1}$ with a polynomial $p\in\mathbb{C}[z]$.
The second case $f(z)=\frac{p(z)}{z^N-1}$ just states that the sequence $(a_n)$ is periodic for large $n$.
For the Euler function we have the formula $$\phi(z) =\prod_{n=1}^{\infty} (1-z^n) =\sum_{n=-\infty}^{\infty}(-1)^n z^{(3n^2-n)/2} $$
The sum has only the coefficients $\{0,\pm 1\}$ and the sequence of coefficients isn't periodic. Therefore, $\mathbb{D}$ is the natural boundary of $\phi(z)$.