Let $X_{n}$ be the maximum score obtained after $n$ throws of a fair dice. Show that $(X_{n})_{n\in\mathbb{N} }$ is a Markov Chain. Let $(X_{n})_{n\in\mathbb{N} }$  be a sequence of randon variables defined by
$$X_{n}:=\mbox{maximum score obtained after } n \mbox{ throws of a fair dice.}$$
I need show that $(X_{n})_{n\in\mathbb{N} }$ is a Markov Chain. In this sense, we have to show that
$$P(X_{n+1}=i_{n+1}|X_{1}=i_{1},X_{2}=i_{2},\ldots,X_{n}=i_{n})=P(X_{n+1}=i_{n+1}|X_{n}=i_{n})\tag{1}$$
I know that if we consider  the random variable  $Y_{k}$ which are the result of $k$th throw of our fair dice, then $(Y_{k})_{k\in\mathbb{N}}$ is iid with uniform distribution over $\left\{1,2,3,4,5,6\right\}$, so, we have $X_{n}=\max\left\{Y_{1},\ldots,Y_{n}\right\}$. Furthermore, the cumulative distribution of $X_{n}$ is
$$F_{X_{n}}(k)=\left\{\begin{array}{ll}
0 & \mathrm{if}\: k<1 \\
\frac{i^{n}}{{6}^{n}} & \mathrm{if}\: i\leq k <i+1 \:\: \mathrm{for}\: i=1,2,3,4,5.\\
1 & \mathrm{if}\:\: k\geq 6
\end{array}\right.$$
Therefore, (1) is equivalent to
$$P(\max\left\{Y_{1},\ldots,Y_{n},Y_{n+1}\right\}=i_{n+1}|Y_{1}=i_{1},\max\left\{Y_{1},Y_{2}\right\}=i_{2},\ldots,\max\left\{Y_{1},\ldots,Y_{n}\right\}=i_{n})=P(\max\left\{Y_{1},\ldots,Y_{n},Y_{n+1}\right\}=i_{n+1}|\max\left\{Y_{1},\ldots,Y_{n}\right\}=i_{n})$$
I have not been able to prove this last, I need help in this matter.
 A: Here is the Markov transition graph for your problem, where the states are labeled by the current maximum and $0$ represents the initial state:

All you need do is calculate the probabilities for the transitions and see that the final state (the sole attractor) is (in probability) $6$.  Each transition depends solely on the current state, not the sequence that led to the current state.   
A: Note that $$X_{n+1}=\max \{Y_1,\dots,Y_{n+1}\} = \max \{\max\{Y_1,\dots,Y_n\},Y_{n+1}\} = \max \{X_n,Y_{n+1}\}.$$
So $X_{n+1}$ only depends on $X_n$ and none of the $X_i$ for $i<n$.
A: Markov Matrix for one-step transition probability would be
\begin{bmatrix}
\frac16 & \frac16 & \frac16 & \frac16 & \frac16 & \frac16 \\
0 & \frac26 & \frac16 & \frac16 & \frac16 & \frac16 \\
0 &  0 & \frac36  & \frac16 & \frac16 & \frac16 \\
0 &  0 & 0  & \frac46 & \frac16 & \frac16 \\
0 &  0 & 0 & 0  & \frac56 & \frac16 \\
0  &  0  &  0  &  0  &  0  &  1 
\end{bmatrix}
The long term limiting probabilities would be as followed
\begin{align}
\pi_6 &= 1 & \pi_1 &= \pi_2 = \pi_3 = \pi_4 = \pi_5 = 0
\end{align}
