Markov Chain Derivation question This page defines the Markov Property as the following.  Does anyone how form (b) becomes (c)?  Is it due to $(X_{n-1} = i_{n-1}) \subset (X_{n-1} = i_{n-1}) \cap ... \cap (X_0 = i_0)$?  Is it due to independence of the conditional events?
a)
$$
\mathbb{P}(X_0 = i_0, X_1 = i_1, X_2 = i_2, ..., X_n = i+n)
$$
b)
$$
= \mathbb{P}(X_0 = i_0) \mathbb{P}(X_1 = i_1 | X_0 = i_0) \mathbb{P}(X_2 = i_2 | X_1 = i_1, X_0 = i_0) ... \mathbb{P}(X_n = i_n | X_{n-1} =i_{n-1} , ..., X_1 = i_1, X_0 = i_0)
$$
c)
$$
= \mathbb{P}(X_0 = i_0) \mathbb{P}(X_1 = i_1 | X_0 = i_0) \mathbb{P}(X_2 = i_2 | X_1 = i_1) ... \mathbb{P}(X_n = i_n | X_{n-1} = i_{n-1})
$$

By the definition of a Markov chain, the probability of the $k$ th state depends only on the $k-1$ th state, it is independent of the $k-2, k-3, \cdots, 0$ th states.
This can be explained best with the help of a state transition diagram.
For example, in this Markov chain, there are no arrows linking state $2$ to state $0$, without passing through state $1$. As the diagram clearly shows, the $k$ th state is linked to the $k-1$ th state only. Hence, $$\mathbb{P}(X_k = i_k|X_{k-1}=i_{k-1}, \ldots, X_0=i_0) = \mathbb{P}(X_k = i_k | X_{k-1} = i_{k-1})$$
