Is there another way to prove this expression over $1/(1-z)$

Question

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 ?

Answer 1

If $|z|<1$, multiply by $1-z$ and FOIL it out.

Answer 2

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).

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\verts{z} < 1}$.