If $|z|<1$ then $\prod_{k=0}^\infty (1+z^{2^k})$ = $(1-z)^{-1}$ Prove that if $|z|<1$ then $\prod_{k=0}^\infty (1+z^{2^k})$ converges and is equal to $(1-z)^{-1}$.  
My attempt:
Note that
\begin{equation}
\begin{aligned}
    (1-z)\prod_{k=0}^N (1+z^{2^k}) &= (1-z)(1+z)(1+z^2)(1+z^{2^2})\cdots(1+z^{2^N})\\
    &= (1-z^2)(1+z^2)(1+z^{2^2})(1+z^{2^3})\cdots(1+z^{2^N})\\
    &= (1-z^{2^2})(1+z^{2^2})(1+z^{2^3})(1+z^{2^4})\cdots(1+z^{2^N})\\
    &= (1-z^{2^{N+1}})
\end{aligned}
\end{equation}
For |z|<1
\begin{equation}
    \lim_{N\to\infty} (1-z)\prod_{k=0}^N (1+z^{2^k})= \lim_{N\to\infty}(1-z^{2^{N+1}}) =1.
\end{equation}
Is this enough to solve the problem? 

You can see by induction on $N$ that
$$
\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}
$$
and passing to the limit, you can deduce that
$$
\prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)=
\sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1
$$
(this last one is the geometric series, which you should know!)
