Help understanding sequence proof Prove $(1 + z + z^{2} + \ldots + z^{9})
\cdot (1 + z^{10} + z^{20} + \ldots + z^{90})
\cdot (1 + z^{100} + z^{200} + \ldots + z^{900})
\cdot \ldots = \frac{1}{1 - z}$.
For $|z|<1$.
De Souza and Silva give the following proof in the book Berkeley Problems in Mathematics:

But I don't understand how they got there. It seems to me several steps are missing.
 A: Note that
$$\prod_{i=0}^{n-1}\left(\sum_{j=0}^9z^{10^ij}\right)=\prod_{i=0}^{n-1}\left(\frac{z^{10^{i+1}}-1}{z^{10^i}-1}\right)=\frac{z^{10^n}-1}{z-1},$$
wich goes to $-1/(z-1)=1/(1-z)$ as $n\to \infty$.
In the first step we've used that
$$
(z^{10^i}-1)\left(1+z^{10^i}+\cdots+z^{9\cdot 10^i}\right) = z^{10^{i+1}}-1,
$$
whence $\sum_{j=0}^9z^{10^i\cdot j}=\frac{z^{10^{i+1}}}{z^{10^i}-1}$, and in the second step, we've used that
$$
\frac{z^{10}-1}{z-1}\cdot \frac{z^{100}-1}{z^{10}-1}\cdots \frac{z^{10^n}-1}{z^{10^{n-1}}-1} = \frac{z^{10^n}-1}{z-1},
$$
because most factors cancel.
A: Alternatively, any integer $m$ has a unique representation in base 10: $m=a_0+a_1\cdot 10+a_2\cdot 10^2+\cdots$. For example $103=3+\cdot 10^0+1\cdot 10^2$, which in exponents would look like $z^{103}=z^{3\cdot 1}\cdot z^{0\cdot 10}\cdot z^{1\cdot 100}$
Convince yourself then that every integer power of $z$ occurs exactly once the infinite product, once expanded. You can formally prove this by expanding a finite product that has sufficiently many terms for a given $z^m$.
