How to prove this infinite product identity?

Question

How can I prove the following identity? $$\large\prod_{k=1}^\infty\frac1{1-2^{1-2k}}=\sum_{m=0}^\infty\left(2^{-\frac{m^2+m}{2}}\prod_{n=1}^\infty\frac{1-2^{-m-n}}{1-2^{-n}}\right)$$ Numerically both sides evaluate to $$2.38423102903137172414989928867839723877...$$

Answer 1

The identity is in fact true for any $0 < x < 1$: $$\prod_{k=1}^{\infty}\frac{1}{1-x^{2k-1}}=\sum_{m=0}^{\infty}\left(x^{\frac{m^2+m}{2}}\prod_{n=1}^{\infty}\frac{1-x^{m+n}}{1-x^{n}}\right) = \sum_{m=0}^\infty\left(x^{\frac{m^2+m}{2}}\prod_{n=1}^{m}\frac{1}{1-x^{n}}\right)$$

We note the telescopic product: $$\displaystyle \prod_{k=1}^{\infty}\frac{1}{1-x^{2k-1}} = \prod_{k=1}^{\infty} \frac{1-x^{2k}}{1-x^k} = \prod_{k=1}^{\infty}(1+x^k)$$

On the other hand the expression:

\begin{align}\prod_{k=1}^{\infty}(1+x^k) &= \sum_{n=0}^{\infty} \left(\sum\limits_{1\le j_1 < j_2 < \cdots < j_n} x^{j_1+\cdots +j_n}\right)\tag{1}\\&= \sum_{n=0}^{\infty} \left(\sum\limits_{1\le k_1,k_2, \cdots ,k_n} x^{nk_1+(n-1)k_2\cdots +k_n}\right)\tag{2}\\&= \sum_{n=0}^{\infty} \left(\sum\limits_{k_1 \ge 1} x^{nk_1}\sum\limits_{k_2 \ge 1}x^{(n-1)k_2} \cdots \sum\limits_{k_n \ge 1}x^{k_n}\right)\tag{3}\\&= \sum\limits_{n=0}^{\infty} \left(\frac{x^n}{1-x^n}\cdot \frac{x^{n-1}}{1-x^{n-1}}\cdots\frac{x}{1-x}\right)\\&= \sum\limits_{n=0}^{\infty} x^{\frac{n^2+n}{2}}\prod\limits_{m=1}^{n}\frac{1}{1-x^m}\end{align}

Justifications:

$(1)$ Coefficient of $z^n:$ in the infinite product $\displaystyle \prod\limits_{k=1}^{\infty}(1+x^kz)$ being $\displaystyle \left(\sum\limits_{1\le j_1 < j_2 < \cdots < j_n} x^{j_1+\cdots +j_n}\right)$.

$(2)$ Made the change of variable $k_m = j_m - j_{m-1}$ for $m \ge 1$ where, $j_0 = 0$.

Then note that $j_1+\cdots +j_n = nk_1+(n-1)k_{2}+\cdots + k_n$

$(3)$ Used the formula for infinite geometric progression: $\displaystyle \sum\limits_{k\ge 1} x^{mk} = \frac{1}{1-x^m}$