Show the following including triple statement How do I show 
\begin{equation*}
\sum \limits_{n=0}^{\infty} z^n=\prod \limits_{m=0}^{\infty}(1+ z^{2^m}) = (1-z)^{-1}?
\end{equation*}
The very left side is obvious because it is the geometric series,but I could not relate the middle one.
 A: It should be
$$\prod_{m\geq 0}(1+z^{2^m}),$$
since:
$$ (1-z)(1+z)(1+z^2)\cdot\ldots\cdot(1+z^{2^N}) = 1-z^{2^{N+1}}.$$
A: Hint. We have
$$
1+z^{2^m}=\frac{1-z^{2^{m+1}}}{1-z^{2^m}}, \quad z\neq1,
$$ then you get a telescoping product. 
A: It's easy to show by induction that 
$$\prod_{k=0}^N(1+z^{2^k})=\sum_{k=0}^{2^{N+1}-1}z^k$$
Base Case:  $N=1$
$$\prod_{k=0}^1(1+z^{2^k})=(1+z)(1+z^2)=1+z+z^2+z^3=\sum_{k=0}^{3}z^k$$
Then, assume there is a number $N$ for which the equality holds.  Now examine the product
$$\begin{align}
\prod_{k=0}^{N+1}(1+z^{2^k})&=\left(1+z^{2^{N+1}}\right)\prod_{k=0}^{N}(1+z^{2^k})\\\\
&=\left(1+z^{2^{N+1}}\right)\sum_{k=0}^{2^{N+1}-1}z^k\\\\
&=\sum_{k=0}^{2^{N+1}-1}z^k+\sum_{k=2^{N+1}}^{2^{N+1}+2^{N+1}-1}z^k\\\\
&=\sum_{k=0}^{2^{N+1}-1}z^k+\sum_{k=2^{N+1}}^{2^{N+2}-1}z^k\\\\
&=\sum_{k=0}^{2^{N+2}-1}z^k
\end{align}$$
which completes the proof.
Now, inasmuch as we have for $|z|<1$, $\sum_{k=0}^{2^{N+1}-1}z^k=\frac{1-z^{2^{N+1}}}{1-z}$, then, 
$$\begin{align}
\lim_{N\to \infty}\prod_{k=0}^N(1+z^{2^k})&=\lim_{N\to \infty}\sum_{k=0}^{2^{N+1}-1}z^k\\\\
&=\lim_{N\to \infty}\frac{1-z^{2^{N+1}}}{1-z}\\\\
&=\frac{1}{1-z}
\end{align}$$
and we are done!
A: The series of partial products begins as:
$$1+x$$
$$(1+x)+(x^2+x^3)$$
$$(1+x+x^2+x^3)+(x^4+x^5+x^6+x^7)$$
$$(1+x+x^2+x^3+x^4+x^5+x^7)+(x^8+x^9+x^{10}+x^{11}+x^{12}+x^{13}+x^{14}+x^{15})$$
and so on. I parenthesized it to emphasize the following point:

When we multiply each term by $(1+x^{2^m})$, we keep the previous terms (the first set of parenthesis), and add a shifted set of terms (the second set of parenthesis).

To formalize what's going on, one can note that it is easy to prove the following inductively:
$$\prod_{m=0}^{k-1}(1+x^{2^m})=\sum_{m=0}^{2^k-1}x^m$$
which clearly implies the infinite series are equal as well.

More intuitively, notice that when we expand the product
$$(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots$$
we get to, at each binomial, choose whether to "take" the $1$ or the $x^{2^m}$. Then we take the product over all of our choices and take the sum over all possible choices - this is just expressing the usual distribution $$(a+b)(c+d)=ac+ad+bc+bd$$
in a new form (see how we choose every pair of either $a$ or $b$ with either $c$ or $d$?). This tells us that each term in the expansion is of the form $x^{2^{a}+2^b+2^c+\ldots}$ where we freely choose $a,b,c,\ldots$ to be the positions where we take the $x^{2^m}$ term. However, realizing that this essentially just says that we have an $x^n$ term for every binary expansion of $n$, it's clear that we have exactly one $x^n$ term for every $n$. 
