Prove/Disprove $\sigma(Z_0, Z_1, ..., Z_n) = \sigma(V_0, V_1, ..., V_n)$ Given a stochastic process $Z = (Z_n)_{n \geq 0}$ on a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F_n}\}, \mathbb{P})$ where $\mathscr{F_n} = \mathscr{F_n}^{Z} \doteq \sigma(Z_0, Z_1, ..., Z_n)$, where
$$P(Z_{n+1} = 2Z_n) = 1/2 = P(Z_{n+1} = 0)$$ and $$Z_0 = 1$$
we can define the independent random variables $V_0 = 1$,
$V_1, V_2, ...$ ~ $P(V_i = 0) = P(V_i = 2) = 1/2$
$\to Z_n = \prod_{i=0}^{n} V_i$.
How do you prove rigorously that $$\sigma(Z_0, Z_1, ..., Z_n) = \sigma(V_0, V_1, ..., V_n)$$? I get it intuitively: we know up to vn iff we know up to zn

What I tried:
$$Z_n = \prod_{i=0}^{n} V_i \to \sigma(Z_n) \subseteq \sigma(V_0, ..., V_n)$$
$$Z_{n-1} = \prod_{i=0}^{n-1} V_i \to \sigma(Z_{n-1}) \subseteq \sigma(V_0, ..., V_{n-1}) \subseteq \sigma(V_0, ..., V_n)$$
$$.$$
$$.$$
$$.$$
$$Z_{0} = \prod_{i=0}^{0} V_i \to \sigma(Z_0) \subseteq \sigma(V_0) \subseteq \sigma(V_0, ..., V_n)$$
So $$\sigma(Z_0), \sigma(Z_1), \dots, \sigma(Z_n) \subseteq \sigma(V_0, ..., V_n)$$
$$\to (\sigma(Z_0) \cup \sigma(Z_1) \cup \dots \cup \sigma(Z_n)) \subseteq \sigma(V_0, ..., V_n)$$
Does this mean
$$\sigma(\sigma(Z_0) \cup \sigma(Z_1) \cup \cdots \cup \sigma(Z_n)) = \sigma(Z_0, Z_1, ..., Z_n) \subseteq \sigma(V_0, ..., V_n)$$ ?
How about the other direction?
$$\sigma(Z_0, Z_1, ..., Z_n) \supseteq \sigma(V_0, ..., V_n)$$ ?
 A: The $\sigma$-algebras are different.
Consider the $n = 2$ case.  
Assume the $Z_i$ ($i = 1,2$) are constructed from the $V_i$ as you wrote in the question.  (I assume the probabilities you wrote for the $Z_i$ are conditional probabilities for $Z_{i+1}$ given $Z_i$, as otherwise your question is inconsistent).
Examine the event $E = \{ V_1 = 0, V_2 = 0 \}$.
(Note on this event we have $Z_0 = 1, Z_1 = 0$.)
By definition of the generated $\sigma$-algebra we have $E \in \sigma(V_1,V_2)$.
The three possible observations for $(Z_1,Z_2)$ are $(0,0)$, $(2,0)$ or $(2,4)$.  
The event $(Z_1,Z_2) = (0,0)$ is $F_1 = \{ V_1= 0, V_2 = 0\} \bigcup \{ V_1= 0, V_2 = 2 \}$.  Note the union here.
The event $(Z_1,Z_2) = (2,0)$ is $F_2 = \{ V_1 = 2, V_2 = 0\}$.
The event $(Z_1,Z_2) = (2,4)$ is $F_2 = \{ V_1 = 2, V_2 = 2\}$.
The smallest $\sigma$ algebra containing these three disjoint events $F_1,F_2,F_3$ can be written down explicitly; it is the power set
$$
\{
\emptyset, F_1, F_2, F_3, F_1\cup F_2,  F_1\cup F_3, F_2\cup F_3, 
F_1\cup F_2 \cup F_3\}.
$$
The event $E$ is not there!
