As noted by others, what you write is NOT the definition of the independence of $(A_1,A_2,\ldots,A_n)$. You should ask that $\mathrm P(A_{k_1}\cap A_{k_2}\cap\cdots \cap A_{k_i})=\mathrm P(A_{k_1})\mathrm P(A_{k_2})\cdots \mathrm P(A_{k_i})$ for every choice of all distinct indices $k_j$ in $\{1,2,\ldots,n\}$.
This point set aside, let me mention that a strategy which might help avoid some tediousness in such a context is to translate everything in terms of random variables. I know, random variables are supposed to be always more complicated than events but in fact, the opposite holds quite often (did somebody just say linearity?), and the present question is a good example of the phenomenon.
We first make the, seemingly odd, general remark that, for every event $A$, $ \mathrm P(A)=\int_\Omega\mathbf 1_A\mathrm dP=\mathrm E(\mathbf 1_A), $ where $\mathbf 1_A$ denotes the indicator function of $A$ defined by $\mathbf 1_A(\omega)=1$ if $\omega\in A$ and $\mathbf 1_A(\omega)=0$ if $\omega\in\Omega\setminus A$.
Turning to the question, let us choose $k$ events from the $n$ events $A_1$, $A_2$, ..., $A_n$, all different, rename them as $B_1$, $B_2$, ..., $B_k$, and introduce $B=\bigcap\limits_{i=1}^k(B_i)^c$. One knows that the indicator function of a complement is $1$ minus the original indicator function and that the indicatior function of an intersection is the product of the indicator functions, hence $ \mathbf 1_B=\prod\limits_{i=1}^k(1-\mathbf 1_{B_i})=Q_k(\mathbf 1_{B_1},\mathbf 1_{B_2},\ldots,\mathbf 1_{B_k}), $ where $Q_k$ denotes the polynomial $ Q_k(x_1,x_2,\ldots,x_k)=\prod\limits_{i=1}^k(1-x_i). $ Like every polynomial, $Q_k(x_1,x_2,\ldots,x_k)$ may be expanded into a sum of monomials in the unknowns $x_1$, $x_2$, ..., $x_k$. Since the (partial) degree of $Q_k$ in each $x_i$ is $1$ the expansion of $Q_k$ involves only monomials of the form $x_{i_1}x_{i_2}\cdots x_{i_\ell}$ for some distinct indices $i_j$. In other words, $ Q_k(x_1,x_2,\ldots,x_k)=\sum_Iq_I\prod_{i\in I}x_i, $ where the sum runs over the $2^k$ subsets $I$ of $\{1,2,\ldots,k\}$, for some coefficients $q_I$ whose values will not be relevant. (The interested reader might note however that $q_\varnothing=1$ and $q_{\{1,2,\ldots,k\}}=(-1)^k$, and the motivated one might show that $q_I=(-1)^{|I|}$ for every $I\subseteq\{1,2,\ldots,k\}$.)
We stress that this relation holds between polynomials, hence every choice of the variables $x_i$ yields an equality, whether these are numbers or functions. In particular, evaluating both sides at the functions $\mathbf 1_{B_i}$ yields $ \mathbf 1_B=\sum\limits_Iq_I\prod\limits_{i\in I}\mathbf 1_{B_i}. $ For every $I$, note that $ \prod\limits_{i\in I}\mathbf 1_{B_i}=\mathbf 1_{B_I},\quad \mbox{where}\ B_I=\bigcap\limits_{i\in I}B_i, $ and that the independence of the events $(B_i)_{i\in I}$ yields $ \mathrm E(1_{B_I})=\mathrm P(B_I)=\prod\limits_{i\in I}\mathrm P(B_i). $ Summing this over every $I$ yields $ \mathrm P(B)=\mathrm E(\mathbf 1_B)=\sum\limits_Iq_I\mathrm P(B_I)=\sum\limits_Iq_I\prod\limits_{i\in I}\mathrm P(B_i)=Q_k(\mathrm P(B_1),\mathrm P(B_2),\ldots,\mathrm P(B_k)), $ where the last equality stems from the very definition of $Q_k$ evaluated at the real numbers $\mathrm P(B_i)$. But one knows the value of $Q_k$ at every point, in particular at $(\mathrm P(B_1),\mathrm P(B_2),\ldots,\mathrm P(B_k))$, which is $ \mathrm P(B)=\prod\limits_{i=1}^k(1-\mathrm P(B_i))=\prod\limits_{i=1}^k\mathrm P((B_i)^c), $ and the proof is over. To conclude, note once again that we used the polynomial $Q_k$ twice, once for functions and the other one for real numbers.