Show lemma on probability of $n$ independent sets 
Let be $(\Omega, \mathcal{A},P)$ a probability space and $A_1,\dots ,A_n \in \mathcal{A}$. For sake of convenience we define $A^1:=A$ and $A^c:=\Omega\setminus A$. Let be $(k_1,\dots , k_n)\in\{1,c\}^n$. Show that
$$
A_1,\dots , A_n\text{ are stochastically independent }\implies P\left(\bigcap\limits_{j=1}^n A_j^{k_j}\right)=\prod\limits_{j=1}^nP(A_j^{k_j}).
$$

My approach:
base case of induction:
Let be $m$ the number of the "complement signs" $c$. If $n=1$, then the equation obviuously holds for each $m$ with $0\leq m\leq 1$.
induction hypothesis:
We assume that the equation holds if we consider $n$-many stocastically independent sets $A_1,\dots A_n$ and $m$-many complement signs, with $0\leq m\leq n$.
induction step $n\to n+1$, with $0\leq m\leq n+1$:
\begin{align*}
P\left(\bigcap\limits_{j=1}^{n+1} A_j^{k_j}\right)=P\left(\bigcap\limits_{j=1}^{n} A_j^{k_j}\cap A_{n+1}^{k_{n+1}}\right)=\dots?
\end{align*}

Now, I am a bit lost because it seems like I should perform another induction on $m$. However, this would require a second induction hypothesis like "the equation holds in the case of $n+1$ and $0\leq m\leq n$" which would beg the question.
Any ideas how to proceed?
 A: The induction on $m$ is sufficient: we show by induction on $m$ that for each number $n \geqslant m$, stochastically independent sets $A_1, \ldots, A_n \in \mathcal{A}$ and $k \in \{ 1, c \}^n$ containing $m$-many complement symbols the equality from the question holds.
For $m=0$ the claim is trivial. Fix $m \geqslant 1$, assume the claim holds for $m-1$ and take $A_1, \ldots, A_n$ and $k \in \{ 1, c \}^n$ as in the statement. By symmetry we can assume that $k_n = c$. We have that
$$\bigcap_{j=1}^{n-1} A_j^{k_j} = \left( \bigcap_{j=1}^{n-1} A_j^{k_j} \cap A_n \right) \cup \left( \bigcap_{j=1}^{n-1} A_j^{k_j} \cap A_n^c \right)$$
and the sets on the right are disjoint, so
$$P \left( \bigcap_{j=1}^{n-1} A_j^{k_j} \right) = P\left( \bigcap_{j=1}^{n-1} A_j^{k_j} \cap A_n \right) + P\left( \bigcap_{j=1}^{n-1} A_j^{k_j} \cap A_n^c \right).$$
The first two probabilities can be computed using the induction hypothesis since among the "exponents" there are exactly $m-1$-many complement signs:
$$\prod_{j=1}^{n-1} P(A_j^{k_j}) = \prod_{j=1}^{n-1} P(A_j^{k_j}) \cdot P(A_n) + P\left( \bigcap_{j=1}^{n-1} A_j^{k_j} \cap A_n^c \right).$$
It follows that
$$P\left( \bigcap_{j=1}^n A_j^{k_j} \right) = P\left( \bigcap_{j=1}^{n-1} A_j^{k_j} \cap A_n^c \right) = \prod_{j=1}^{n-1} P(A_j^{k_j}) \cdot \big(1-P(A_n)\big) = \prod_{j=1}^n P(A_j^{k_j}),$$
as desired.
