Gauss measure is not a pushforward of product measure Let $ N = \{1,2,3, \ldots \}$. We define $\varphi : N^N \mapsto [0,1]$ as 
$$ \varphi \left( (a_n)_{n \in_N} \right) = [0;a_1, a_2, \ldots ]$$
Where the expression on the right is a infinite continued fraction:
$$[0;a_1,a_2,a_3,\dotsc] = \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}}$$
 It is standard exercise that this is borel mapping (continuous in fact, see here).
We define the Gauss measure on $[0,1]$ as:
$$ d\nu = \frac{1}{\ln(2)}\,\frac{dx}{1+x} $$
Let $\mu$ be a probability measure on $N$. By a product measure on $N^N$ i mean probabilistic measure $\widetilde { \mu }$ such that for any set of the form $F = F_1 \times \cdots \times F_n \times N \times N \times \cdots$ holds equality 
$$ \widetilde {\mu } (F) = \prod_{n=1}^n \mu (F_k) $$
Such a measure exists and is unique due to teorem about extending funtion on ring to a measure.
I am to prove that the Gauss measure $\nu$ is not a pushforward of any such measure $\mu$, meaning that for any $\mu$ the equality 
$$ \nu( A ) = \mu ( \varphi^{-1} [A])$$
cannot hold for every borel $A \subset [0,1]$.
I am looking for any help. I have no experience with infinite measure products. This is an exercise for my ergodic theory course.

I had obtained a tip that i don't understand:

Product measure has the property that basis clopen sets with disjoint supports are stochastically independent. Gauss measure does not have this property.

 A: Thanks to @Jonas i came up with a solution and my supervisor accepted it. Hence, i am posting it for completeness. Don't judge me if it is poorly written, i am translating it from Polish. ;)
From the source given in question we know that for any irrational  $x \in [0,1]$ it correspond to some continued fraction $[0; a_1, a_2, \ldots ]$. Another thing that we know is that it is between points
$$ x_n = [ 0; a_1, a_2, \ldots , a_n] \qquad x_{n+1} = [0; a_1, a_2, \ldots, a_{n+1} ] $$
To ease the notation lets denote for any $ \sigma \in N^{ < N}$ the set $ [[ \sigma ]] = \{ f \in N^N : \sigma \prec f \}$ and by $| \sigma|$ the length of $\sigma$.
Assume that for some product measure $\widetilde{ \mu}$ we have for all borel sets $A$ 
$$ \nu (A) = \widetilde{ \mu} \left( \varphi^{-1} (A) \right) $$
From the fact that $\varphi$ is a bijection  from $N^N$ to  $[0,1] \setminus Q$ and easy-to-check equalities
$$[0;1] = 1 \qquad [0;2] = 1/2 \qquad [0;3] = 1/3 \qquad [0;1,2] = 2/3 \qquad [0;1,3] = 3/4 $$
we get the following equalities:
$$ \varphi^{-1} [1/2, 1] = [[(1) ]] \qquad \varphi^{-1} [1/3 ,1/2] = [[(2) ]] \qquad \varphi^{-1} [2/3, 3/4] = [[(1,2) ]] $$
Hence, from the definition of product measure we have that
$$ \widetilde{\mu} \left( [[(1,2)]] \right)= \widetilde{\mu} \left( [[(1)]] \right) \cdot \widetilde{\mu} \left( [[(2)]] \right)$$
By assertion that $\nu$ is a pushforward measure we should have $ \nu (2/3, 3/4) = \nu (1/2,1) \nu (1/3, 1/2)$. However, with the help of almighty Wolfram we can tell that this is not the case because:
$$\nu (2/3, 3/4) = (\log 2)^{-1} ( \log 21 -\log 20) \qquad \nu (1/2,1) =  (\log 2)^{-1} ( \log 4 -\log 3) \qquad  \nu (1/3, 1/2) = (\log 2)^{-1} ( \log 9 -\log 8)$$
and
$$( \log 4- \log 3)(\log 9 - \log 8) \neq \log 2 ( \log 21 - \log 20) $$
So we have contradiction.
