Given a Markov Chain $X_n$ such that $P(X_{n+1}|X_n,...X_0) = P(X_{n+1}|X_n)$. Show that
$$P(X_0|X_1,...,X_n) = P(X_0|X_1)$$
My approach was to start like $$P(X_0|X_1,...,X_n) = \frac{P(X_0,X_1,...,X_n)}{P(X_1,...,X_n)} $$
Then I think I need to condition on something and the numerator and denominator will cancel out leaving my desired result, but so far anything I thought to condition on has not worked out.
You can prove this by induction via
\begin{eqnarray} P(X_0\mid X_1,\ldots,X_n) &=& \frac{P(X_0,\ldots,X_n)}{P(X_1,\ldots,X_n)} \\ &=&\frac{P(X_n\mid X_0,\ldots,X_{n-1})P(X_0,\ldots,X_{n-1})}{P(X_n\mid X_1,\ldots,X_{n-1})P(X_1,\ldots,X_{n-1})} \\ &=&\frac{P(X_n\mid X_{n-1})P(X_0,\ldots,X_{n-1})}{P(X_n\mid X_{n-1})P(X_1,\ldots,X_{n-1})} \\ &=&\frac{P(X_0,\ldots,X_{n-1})}{P(X_1,\ldots,X_{n-1})} \\ &=& P(X_0\mid X_1,\ldots,X_{n-1})\;. \end{eqnarray}