Prove complements of independent events are independent. Given a finite set of events $\{A_i\}$ which are mutually independent, i.e., for every subset $\{A_n\}$, 
$$\mathrm{P}\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathrm{P}(A_i).$$
show that the set $\{A_i^c\}$, that is the set of complements of the original events, is also mutually independent.
I can prove this, but my proof relies on the Inclusion-Exclusion principle (as does the proof given in this question). I'm hoping there is a more concise proof.
Can this statement be proved without the use of the Inclusion-Exclusion principle?
 A: Let $[n] = \{1, \cdots, n\}$ and $p(X_1, \cdots, X_n) = \sum_{I \subseteq [n] } a_{I} \prod_{i \in I} X_i$. That is, $p$ is a polynomial in $X_1, \cdots, X_n$ consisting of square-free monomials. If $\{A_1, \cdots, A_n\}$ are mutually independent, then
\begin{align*}
\Bbb{E}[p(\mathbf{1}_{A_1}, \cdots, \mathbf{1}_{A_n})]
&= \sum_{I \subseteq [n] } a_{I} \Bbb{E}\Big( \prod_{i \in I} \mathbf{1}_{A_i} \Big) \\
&= \sum_{I \subseteq [n] } a_{I} \Bbb{P}\Big( \bigcap_{i \in I} {A_i} \Big) \\
&= \sum_{I \subseteq [n] } a_{I} \prod_{i \in I} \Bbb{P} (A_i) \\
&= p(\Bbb{P}(A_1), \cdots, \Bbb{P}(A_n)),
\end{align*}
Then the conclusion follows for the choice $p(X_1, \cdots, X_n) = (1-X_1)\cdots(1-X_n)$. 
A: Hint: prove that the set of events stays independent if you replace one of them by its complement, i.e. that given your conditions the set $\{A_1^c, A_2, \ldots, A_n\}$ is independent. Then use this $n$ times to replace all of $A_i$ by their complements one by one.
Update. Hint 2: to avoid clutter, let me show you what I mean on the example of two events, $B$ and $C$. Suppose $B$ and $C$ are independent, i.e. $P(B \cap C) = P(B) \cdot P(C)$. We want to show that $B^c$ and $C$ are independent. Indeed:
$$
\begin{align*}
P(B^c \cap C) &= P(C \setminus (B \cap C)) \\
&= P(C) - P(B \cap C) \\
&= P(C) - P(B) \cdot P(C) \\
&= P(C) \cdot (1-P(B)) \\
&= P(B^c) \cdot P(C).
\end{align*}
$$
I didn't use inclusion-exclusion here. And this approach scales, i.e. it works the same if you consider more than $2$ variables.
