# Jacobi Triple Product using Euler's identities

I recently came across a simple proof for Jacobi's triple product (here), in the proof Andrews assumes two identities:

$$E_1 = \prod_{n=0}^{\infty}(1+x^nz) = \sum_{n=0}^{\infty}\frac{x^{n(n-1)/2}z^n}{(1-x)...(1-x^n)}$$ and $$E_2 = \prod_{n=0}^{\infty}(1+x^nz)^{-1} = \sum_{n=0}^{\infty}\frac{(-1)^nz^n}{(1-x)...(1-x^n)}$$

While the proof itself is well written and easy to understand, I can't find a proof of the two identities given in the paper, mostly because their names is not stated and the reference is from a book I can't find online.

Would anyone be able to give a short proof or redirect me to a proof of those two identities?

Thank you

These identities are pretty standard. Let's write $$F(x, z)=\prod_{n=0}^{\infty} (1+x^nz)=\sum_{n=0}^{\infty}a_nz^n\tag{1}$$ We can see that $$F$$ satisfies functional equation $$F(x, z) =(1+z) F(x, zx)\tag{2}$$ and using the infinite series for $$F$$ in above equation we get $$\sum_{n=0}^{\infty}a_nz^n=(1+z)\sum_{n=0}^{\infty}a_nx^nz^n$$ Equating coefficients of $$z^n$$ on both sides we get $$a_n=a_nx^n+a_{n-1}x^{n-1}$$ or $$a_n=\frac{x^{n-1}}{1-x^n}a_{n-1}$$ Using this equation recursively and noting that $$a_0=1$$ we get $$a_n=\frac{x^{n(n-1)/2}}{(1-x)(1-x^2)\dots(1-x^n)}$$ and the first identity in question is proved. The second identity can be established in similar manner.