Q:

Showing that if the conditional probabilities exist.

I'm trying to show that
If the conditional probabilities exist, then
 P(A1 $\cap$ A2 $\cap$ ... $\cap$An) = P(A1)P(A2|A1)...P(An|A1$\cap$A2$\cap$...$\cap$An-1).
I'm unsure how to approach these sorts of problems as I'm still fairly new to Probability.
I started off with saying:
Let A1, A2, ..., An be events, each with P(Ai) >= 0 for all i $\in$ {1, 2, 3, ...}.
We have by our hypothesis that the conditional probabilities exist, so by definition we have that:
P(A2|A1) = $\frac{P(A2 \cap A1)}{P(A1)}$
...
P(An|An-1) = $\frac{P(An \cap An-1)}{P(An-1)}$
Then by the multiplication law:
P(An-1 $\cap$ An) = P(An|An-1)P(An-1)
(and so on).
So
P(A1 $\cap$ A2 $\cap$ ... $\cap$An) = P(A1 $\cap$ A2) P(A3 $\cap$ A4) ... P(An-1$\cap$An)
Before I even go on, is what I did here correct so far? The above line strikes me as not convincing as I'm not sure if there is such a property to allow me to do what I did above. The idea I had was to substitute an n number of conditional probability equations (in the form of the multiplicative law) and then multiply them all together.
But generally speaking I am not sure if I understand how to approach these sorts of problems - so any help appreciated.
Do note that this is elementary probability so I only have a limited tool set I can use.

A:

note that the k'th term is
$$P(A_k| A_1\cap A_2\cap ...\cap A_{k-1} ) = \frac{P( A_1\cap A_2\cap ...\cap A_{k-1}\cap A_k )}{ P(A_1\cap A_2\cap ...\cap A_{k-1} )}$$
from which you obtain a telescopic product

