Show markov property for a s.p. with independent increment I am reading a textbook and get confused on one example which shows $X_t$ with independent increments and $X_0=0$ is a Markov process. In the proof\begin{align*}\mathbb{P}(X_t\leq x\mid X_{t_1},\ldots ,X_{t_n}) 
& =\mathbb{P}(X_t-X_{t_n}+X_{t_n}\leq x\mid X_{t_1},\ldots ,X_{t_n}) \\
& =\mathbb{P}(X_t-X_{t_n}+X_{t_n}\leq x\mid X_{t_n}) \\
& =\mathbb{P}(X_t\leq x\mid X_{t_n})
\end{align*}
I can understand that since the increments $X_t-X_{t_n}$ are independent, so we can remove $X_{t_1},\ldots ,X_{t_{n-1}}$ from the condition, but how to deal with the $X_{t_n}$? Is it also independent on $X_{t_1},\ldots ,X_{t_{n-1}}$?

After doing some small manipulations about the idea of independent increment, I think I have figured out a method:
\begin{align*}
\mathbb{P}(X_t \leq x | X_{t_1},\ldots ,X_{t_n}) 
&= \mathbb{P}(X_t-X_{t_n} \leq x-X_{t_n} | X_{t_1},\ldots ,X_{t_n}) \\
&= \mathbb{P}(X_t-X_{t_n} \leq x-X_{t_n} |X_{t_n}-X_{t_{n-1}}, X_{t_{n-1}}- X_{t_{n-2}}, \ldots, X_{t_2} - X_{t_1}) \\
&= \mathbb{P}(X_t-X_{t_n} \leq x-X_{t_n} |X_{t_{n}}) \\
&= \mathbb{P}(X_t \leq x | X_{t_{n}})
\end{align*}
