Can someone please tell me, how from these lecture notes, on pp. 6, equation (1) is derived (where the joint probabilities $$P(X_0=i_0,\ldots,X_n=i_n)$$of a Markov chain $(X_n)_{n\in \mathbb{N}}$ is computed)?
I have to somehow use Lemma 1 but I can't figure out how.
Note that $P(X_0 = i_0 , \ldots, X_n = i_n) = P(X_n = i_n | X_{n-1} = i_{n-1}, \ldots, X_0 = i_0) P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)$ by the definition of conditional probability. Then, by the markov property, $P(X_n = i_n | X_{n-1} = i_{n-1}, \ldots, X_0 = i_0) = P(X_n = i_n | X_{n-1} = i_{n-1})$.
So, $P(X_0 = i_0 , \ldots, X_n = i_n) = P(X_n = i_n | X_{n-1} = i_{n-1}) P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)$.
Now, repeat this idea on $P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)$ to end up with the desired statement: $P(X_0 = i_0 , \ldots, X_n = i_n) = P(X_n = i_n | X_{n-1} = i_{n-1}) P(X_{n-1} = i_{n-1} | X_{n-2} = i_{n-2})\ldots P(X_1 = i_1 | X_{0} = i_{0}) P(X_0 = i_0)$.
