<s>[INST] 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.
 [/INST] 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!
</s>