Suppose that we are going to prove that a stochastic process $\{X_n\}$ is a Markov Chain. We have to show that: $$Pr(X_k=i_k |X_{k-1}=i_{k-1},X_{k-2}=i_{k-2},...,X_{0}=i_{0}) = Pr(X_k=i_k|X_{k-1}=i_{k-1})$$
We are using induction method and suppose we have shown that the equation above holds for k = 2. We assume that it holds for k = n-2 as well. Now we have to show that it holds for k=n to complete the proof.
$$=Pr(X_n=i_n |X_{n-1}=i_{n-1},X_{n-2}=i_{n-2},...,X_{0}=i_{0}) \\ = \frac{Pr(X_n=i_n,X_{n-1}=i_{n-1},X_{n-2}=i_{n-2},...,X_{0}=i_{0})}{Pr(X_{n-1}=i_{n-1},X_{n-2}=i_{n-2},...,X_{0}=i_{0}) }\\ = \frac{Pr(X_n=i_n,X_{n-1}=i_{n-1},X_{n-2}=i_{n-2}|X_{n-3}=i_{n-3},...,X_{0}=i_{0})}{Pr(X_{n-1}=i_{n-1},X_{n-2}=i_{n-2}|X_{n-3}=i_{n-3},...,X_{0}=i_{0}) }\\$$
By assumption this is true:
$$Pr(X_{n-2}=i_{n-2} |X_{n-3}=i_{n-3},...,X_{0}=i_{0}) = Pr(X_{n-2}=i_{n-2} |X_{n-3}=i_{n-3})$$
Question here is that can we do this transformation:
$$= Pr(X_{n-1}=i_{n-1},X_{n-2}=i_{n-2}|X_{n-3}=i_{n-3},...,X_{0}=i_{0})\\ = Pr(X_{n-1}=i_{n-1},X_{n-2}=i_{n-2}|X_{n-3}=i_{n-3}) $$