Hey fex.

Since we know Z_n is I.I.D then this implies P(Z_n+1|Z_n,Z_n-1,...) = P(Z_n+1|anything) = P(Z_n+1) (complete independence).

So now we have a distribution involving X_n and Z_n but since Z_n is I.I.D then the probability for f(Xn.Zn) will be P(Xn)*P(Zn).

With this factorization you should be able to do some re-arranging to get P(Xn+1|Xn,Xn-1,Xn-2,...) = P(Xn+1|Xn).