Theorem on intersection of events using conditional probability I understand that
$$P(A,B)=P(B)P(A\mid B)=P(A)P(B\mid A)$$
but the generalization of it is a bit confusing to me.
$$P(A_1,A_2,\ldots,A_n)=P(A_1)P(A_2\mid A_1)P(A_3\mid A_2,A_1)\cdots P(A_n\mid A_1,\ldots,A_{n-1})$$
What I would expect is:
$$P(A_1,A_2,\ldots,A_n)=P(A_n\mid A_1,\ldots,A_{n-1})$$
Please explain why it is not so. Venn diagrams are greatly appreciated. 
 A: From the identity $\Pr(A,B) = \Pr(B)\Pr(A\mid B)$ one gets this:
$$
\Pr(A_1,A_2,\ldots,A_n) = \Pr(A_1,\ldots,A_{n-1}) \Pr(A_n\mid A_1,\ldots,A_{n-1}).
$$
Then by applying that first identity again, one gets this:
\begin{align}
& \Pr(A_1,A_2,\ldots,A_n) \\[8pt]
= {} & \Pr(A_1,\ldots,A_{n-1}) \Pr(A_n\mid A_1,\ldots,A_{n-1}) \\[8pt]
= {} & \Pr(A_1,\ldots,A_{n-2}) \Pr(A_{n-1} \mid A_1,\ldots,A_{n-2})\Pr(A_n\mid A_1, \ldots, A_{n-1})
\end{align}
and so on.  Keep going until you've got it.  (Or more formally: This is mathematical induction.)
A: It cannot be so since setting $A=A_1,A_2,\ldots,A_{n-1}$, $B=A_n$ yields $$\Pr(A,B)=\Pr(B\mid A)$$ which is certainly wrong.
Now if you take the correct formula and take a sequence of nested sequences $$\bar{A}_k = A_1, A_2,\ldots, A_k, k\in\{1,\ldots,n-1\}$$
that is
$$\bar{A}_{n-1} \supset \bar{A}_{n-2}\supset \ldots\supset \bar{A}_2\supset \bar{A}_{1}.$$
Then,
\begin{eqnarray*}
  \Pr(A_1,\ldots,A_n)=\Pr(A_n \mid \bar{A}_{n-1})\Pr(\bar{A}_{n-1})&&\\
\Pr(\bar{A}_{n-1})&=\Pr(A_{n-1}\mid\bar{A}_{n-2})\Pr{(\bar{A}_{n-2})}&&\\
&&\vdots\\
&&\Pr(\bar{A}_2)=\Pr(A_2\mid \bar{A}_1)\Pr{(\bar{A}_1)}=\Pr(A_2\mid{\bar{A}}_1)\Pr{(A_1)}
\end{eqnarray*}
Which leads to the generalized formula.
