I have a discrete-time process that satisfies for each $k\in\mathbb{N}$: $$\forall n_0<n_1<n_2<\ldots<n_k<n\in \mathbb{Z}: \ \ E[X[n+1]|X[n],X[n_k],X[n_{k-1}],\ldots , X[n_0]]=X[n]$$ I want to show that $X[n]$ is a martingale, i.e. that it satisfies: $$\forall n_0<n_1<n_2<\ldots <n_k<n\in \mathbb{Z}:\ \ E[X[n+1]|X[n_k],X[n_{k-1}],\ldots,X[n_0]]=X[n_k]$$.
So I am suggested to prove it by induction on $m$ where $n_k=n-m$. The base step of the induction is easily satisfied since for $m=0$ we get from the definition of $X[n]$ that: $E[X[n+1]|X[n_k],X[n_{k-1}],\ldots, X[n_0]]=E[X[n+1]|X[n_k],X[n_k],X[n_{k-1}],\ldots, X[n_0]]=X[n_k]$.
But I don't see how to prove the induction step: Suppose it's satisfied for $m$ and prove it for $m+1$, i.e: $$E[X[n+1]|X[n-m],X[n_{k-1}],\ldots, X[n_0]]=X[n_k]=X[n-m]$$ Now I need to show that: $E[X[n+1]|X[n-m-1],\ldots]=X[n-m-1]$, but I don't see how exactly.
Can you help me with this derivation?