Let's say I apply Bayes law to the joint CDF of an IID random process that is a Markov chain, and I get something like this:
$$F_X(x_1,~ x_2,~ \cdots,~ x_n) = F_X(x_n~ |~ x_{n-1})~F_X(x_1,~ x_2,~ \cdots,~ x_{n-1})\tag{1}$$
I'm suppose to use this relation to prove that the following law holds for a Markov chain:
$$F_X(x_1,~ x_2,~ \cdots,~ x_n) =F_X(x_1)~F_X(x_2 | x_1)~ F_X(x_3| x_2)\cdots F_X(x_n| x_{n-1})$$
$$F_X(x_1,~ x_2,~ \cdots,~ x_n) = F_X(x_1)~\prod \limits_{k=2}^{n} F_X(x_k~ |~ x_{k-1})\tag{2}$$
By repeating applying this rule (1)... which doesn't seem to give me the required equation to get (2) ...also, when they say its second order distribution... is this the distribution they are referring to: $F_X(x_k~ |~ x_{k-1})$?
$$F_X(x_1,~ x_2,~ \cdots,~ x_n) = F_X(x_n~ |~ x_{n-1})~F_X(x_1,~ x_2,~ \cdots,~ x_{n-1})$$
$$F_X(x_1,~ x_2,~ \cdots,~ x_{n-1}) = F_X(x_{n-1}~ |~ x_{n-2})~F_X(x_1,~ x_2,~ \cdots,~ x_{n-2})$$
$$F_X(x_1,~ x_2,~ \cdots,~ x_{n-2}) = F_X(x_{n-2}~ |~ x_{n-3})~F_X(x_1,~ x_2,~ \cdots,~ x_{n-2})$$
combining these together we have:
$$\begin{aligned}F_X(x_1,~ x_2,~ \cdots,~ x_n) =& F_X(x_n~ |~ x_{n-1})\times \\&F_X(x_{n-1}~ |~ x_{n-2})\times \\&F_X(x_1,~ x_2,~ \cdots,~ x_{n-2})\end{aligned}$$
$$\begin{aligned}F_X(x_1,~ x_2,~ \cdots,~ x_n) =& F_X(x_n~ |~ x_{n-1})\times\\&F_X(x_{n-1}~ |~ x_{n-2})\times\\ ~~~&F_X(x_{n-2}~ |~ x_{n-3})\times\\~~~&F_X(x_1,~ x_2,~ \cdots,~ x_{n-3})\end{aligned}$$
Which can be repeated recursively until the joint probability on the RHS becomes just $F_X(x_1)$ Thus.
$$\begin{aligned}F_X(x_1,~ x_2,~ \cdots,~ x_n) =& F_X(x_n~ |~ x_{n-1})\times\\&F_X(x_{n-1}~ |~ x_{n-2})\times \\ &F_X(x_{n-2}~ |~ x_{n-3}) ~\times\\ &\cdots \\~~~& F_X(x_{2}~ |~ x_{1})~~\times \\& F_X(x_1)\end{aligned}$$
$$\boxed{\small F_X(x_1,~ x_2,~ \cdots,~ x_n) = \bigg(\prod \limits_{i=2}^{n}F_X(x_n~ |~ x_{n-1})\bigg) F_X(x_1)}$$