For any finite $S\subset\left\{ 0,1,2,\ldots\right\} $ define $Z_{S}:=\prod_{s\in S}X_{s}$. Note that here $Z_{n}$ and $Z_{\left\{ 0,\ldots,n\right\} }$
denote the same random variable.
By induction it can be shown that $Z_{n}$ has exactly the same distribution
as - let's pick one out - $X_{0}$ . It uses the apparant independency
of $Z_{n-1}$ and $X_{n}$. Note that $Z_{n}$ can only take values
in $\left\{ -1,1\right\} $ so that it is enough to prove that $P\left\{ Z_{n}=1\right\} =\dfrac{1}{2}$.
$P\left\{ Z_{n}=1\right\} =P\left\{ Z_{n-1}=1\wedge X_{n}=1\right\} +P\left\{ Z_{n-1}=-1\wedge X_{n}=-1\right\} =P\left\{ Z_{n-1}=1\right\} P\left\{ X_{n}=1\right\} +P\left\{ Z_{n-1}=-1\right\} P\left\{ X_{n}=-1\right\} =\dfrac{1}{2}\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{2}$.
In fact this means that $Z_{S}$ has the same distribution as $X_{0}$ for
any finite $S\subset\left\{ 0,1,2,\ldots\right\} $.
Let $n,k$ be integers with $k>0$ and let $S:=\left\{ n+1,\cdots,n+k\right\} $.
Then $Z_{n}$ and $Z_{S}$ are independent and for $\varepsilon,\delta\in\left\{ -1,1\right\} $
we find:
$P\left\{ Z_{n}=\varepsilon\wedge Z_{n+k}=\delta\right\} =P\left\{ Z_{n}=\varepsilon\wedge Z_{S}=\delta\varepsilon^{-1}\right\} =P\left\{ Z_{n}=\varepsilon\right\} P\left\{ Z_{S}=\delta\varepsilon^{-1}\right\} =\dfrac{1}{2}\times\dfrac{1}{2}=P\left\{ Z_{n}=\varepsilon\right\} P\left\{ Z_{n+k}=\delta\right\} $.
This proves that $Z_n$ and $Z_{n+k}$ are independent.