I want to prove $a_0(q)= \prod_{n=1}^{\infty} (1-q^{2n})$ fromo following relation \begin{align} \frac{a_0(q)}{a_0(q^4)} = \prod_{n=1}^{\infty} \frac{(1-q^{2n})}{(1-q^{8n})} \end{align} with inital condition $a_0(0)=1$.
Now from the cross ratio \begin{align} \frac{a_0(q)}{\prod_{n=1}^{\infty}(1-q^{2n})} = \frac{a_0(q^4)}{\prod_{n=1}^{\infty}(1-q^{8n})} = \cdots = \frac{a_0(q^{4^k})}{\prod_{n=1}^{\infty}(1-q^{2\times 4^k n})} \end{align}
The textbook says this shows $a_0(q) = \prod_{n=1}^{\infty} (1-q^{2n})$ but I have no idea why this is true.
For integers $k\ge0$ and $|q|<1$, $$\frac{a_0(q)}{a_0\left(q^{4^{k+1}}\right)}=\prod_{j=0}^k\frac{a_0\left(q^{4^j}\right)}{a_0\left(q^{4^{j+1}}\right)}=\prod_{j=0}^k\prod_{n\ge1}\frac{1-q^{n2^{2j+1}}}{1-q^{n2^{2j+3}}}=\prod_{n\ge1}\prod_{j=0}^k\frac{1-q^{n2^{2j+1}}}{1-q^{n2^{2j+3}}}=\prod_{n\ge1}\frac{1-q^{2n}}{1-q^{2^{2k+3}n}}.$$I think you also need to assume continuity in a neighbourhood of $q=0$. Taking $k\to\infty$,$$\frac{a_0(q)}{a_0(0)}=\prod_{n\ge1}(1-q^{2n}).$$