$M_{n}(z) = \sum_{i=1}^{10}z^{10^{n}(i-1)}$
$\quad \quad =z^{0*10^{n}}+z^{1*10^{n}}+z^{2*10^{n}}+\ldots +z^{9*10^{n}}$
$\quad \quad =1+z^{1*10^{n}}+z^{2*10^{n}}+\ldots +z^{9*10^{n}}$
$\quad\quad = \dfrac{1-(z^{10^{n}})^{10}}{1-z^{10^{n}}}$
$\quad \quad = \dfrac{1-z^{10^{(n+1)}}}{1-z^{10^{n}}}$
$M_{n}(z)$ $= \dfrac{1-z^{10^{(n+1)}}}{1-z^{10^{n}}}$
$Now, $
$\prod_{i=0}^{\infty}M_{i}(z)= M_{0}(z){\times}M_{1}(z){\times}M_{2}(z){\times}\ldots $
$\quad = \left(\dfrac{1-z^{10^{1}}}{1-z^{10^{0}}}\right){\times}\left(\dfrac{1-z^{10^{2}}}{1-z^{10^{1}}} \right ){\times}\left(\dfrac{1-z^{10^{3}}}{1-z^{10^{2}}} \right ){\times}\ldots {\times}\left(\dfrac{1-z^{10^{k}}}{1-z^{10^{(k-1)}}} \right ){\times}\left(\dfrac{1-z^{10^{(k+1)}}}{1-z^{10^{k}}} \right ){\times}\ldots$
$\quad = \dfrac{1}{1-z}$
for ending terms, as $ |z|<1, z^{\infty}$ tends to $0,$ $1-z^{\infty} $ tends to $1.$
Correct Answer: $B$