Can $\prod_{n=1}^\infty(1+e^{2\pi i n\tau})$ be expressed in terms of the Euler Phi Function?

Question

I am wondering if $$(-e^{2\pi i \tau},e^{2\pi i \tau})_\infty=\prod_{n=1}^\infty(1+e^{2\pi i n\tau})$$ can be expressed in terms of Euler's Phi Function. Any help is appreciated.

Answer 1

Observe that $$1+e^{2\pi i n\tau}=\frac{1-(e^{4\pi i\tau})^n}{1-(e^{2\pi i\tau})^n}$$ (this is just $1+a=\frac{1-a^2}{1-a}$). Hence $$\prod_{n=1}^\infty(1+e^{2\pi i n\tau})=\frac{\prod_{n=1}^\infty(1-(e^{4\pi i\tau})^n)}{\prod_{n=1}^\infty(1-(e^{2\pi i\tau})^n)}=\frac{\phi(e^{4\pi i\tau})}{\phi(e^{2\pi i\tau})}=\frac{\phi(q^2)}{\phi(q)}$$ for $q=e^{2\pi i\tau}$.