Suppose $\F_n$ is a filtration and $X_n:\O\rar S$ are measurable with respect to $\F_n$. Assume that there is some $m\in\N$ such that for all bounded measurable $f:S\rar\R$ and all $n$:
$$
\E(f(X_{n+1})|\F_n)
=\E(f(X_{n+1})|\s(X_n,\ldots,X_{n-m+1}))~.
$$
Then $X_n$ is called an $m$-Markov chain. Verify that $Z_n\colon=(X_n,\ldots,X_{n-m+1})$ is a Markov chain with respect to $\F_n$ in $S^m$.
Put $f(x_1,\ldots,x_m)=f_1(x_1)\cdots f_m(x_m)$, then
\begin{eqnarray*}
\E(f(Z_{n+1})|\F_n)
&=&\E(f_m(X_{n+1})\cdots f_1(X_{n-m+1})|\F_n)\\
&=&\E(f_m(X_{n+1})|\s(X_n,\ldots,X_{n-m+1})))f_{m-1}(X_n)\cdots f_1(X_{n-m+1})
\end{eqnarray*}
i.e. $\E(f(Z_{n+1})|\F_n)$ is $\s(Z_n)$-measurable, i.e.
$\E(f(Z_{n+1})|\F_n)=\E(f(Z_{n+1})|Z_n)$. The assertion follows by the Monotone Class Theorem.