Is there another way to prove this expression over $1/(1-z)$ I came across the following relationship:
$$
\frac{1}{1-z} = (1+z)(1+z^2)(1+z^4)(1+z^8)...
$$
If induction is used, the statement can be proven given that:
$$
(1+z)(1+z^2)=1+z+z^2+z^3
$$
and
$$
(1+z)(1+z^2)(1+z^4)=1+z+z^2+z^3+z^4+z^5+z^6+z^7
$$
and so on and so forth ...
Since:
$$
\frac{1}{1-z}=\sum_k z^k
$$
The relationship follows... However, am wondering, is there another way to prove the first equation aside from using induction ?
 A: The coefficient of $x^k$ in $\prod_{n=0}^\infty (1+x^{2^n})$ is the number of ways of writing $k$ as a sum of distinct powers of $2$. There is a unique such way of writing any $k\geq 0$ (this is essentially just a base $2$ expression for $k$, which is unique).
A: If $|z|<1$, multiply by $1-z$ and FOIL it out.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&1 - z^{2^{n}} = \pars{1 - z^{2^{n - 1}}}\pars{1 + z^{2^{n - 1}}} =
\pars{1 - z^{2^{n - 2}}}\pars{1 + z^{2^{n - 2}}}\pars{1 + z^{2^{n - 1}}}
\\[5mm] &\
= \cdots =\pars{1 - z}\pars{1 + z}\ldots\pars{1 + z^{2^{n - 1}}}
\end{align}
Takes the limit $\ds{n \to \infty}$ with $\ds{\verts{z} < 1}$.
