According to calculations in this paper https://www.stat.cmu.edu/~cshalizi/462/lectures/06/markov-mle.pdf
The first line just uses the definition of conditional probability, but the second line actually uses the Markov property — that the future is independent of the past, given the present.
Then the equation (4) becomes equation (5). I did not understand the logic of using $N_{ij}$ here. If I assume that $n=3$ ,then the equation (4) is
$$Pr(X_1,X_2,X_3) = Pr(X_1=x_1)Pr(x_2|x_1)Pr(x_3|x_2) $$
Even if we use the $N_{ij}$ , it will always be equal to the $1$.So why do we need to use $N_{ij}$?
Thank you
You can see that by taking an example, say a MC with just $k=2$ states $\{1,2\}$. Then you run it for $n=10$-steps. Then the formula (4) is still valid but note that each $p_{ij}$ can not just occur once. The total number of transitions will be $9$ (assuming the initial time point is $1$). Therefore some of those $p_{ij}$'s must repeat and how many times? Precisely the number of times the MC jumps from $i$ to $j$.
More precisely, equation (3) or (4) are probabilities of a path visited by the MC. Take for example (in the $2$-states case) a path $(x_1=1,x_2=1,x_3=2,x_4=1,x_5=1,x_6=2,x_7=2,x_8=1,x_9=1,x_{10}=2)$. Then using equation (3) or (4) gives probability of this path is equal to $$ Pr(X_1=1)p_{11}p_{12}p_{12}p_{21}p_{11}\ldots p_{12}= Pr(X_1=1)p_{11}^{3}p_{12}^{3}p_{21}^{2}p_{22}. $$