Consider the canonical Markov Chain in $\mathbb{Z}^d$ with transition kernel $$K(x,dy)=\frac{1}{2d} \sum_{e \in \mathbb{Z}^d,|e|=1} \delta_{x+e}(dy)$$ corresponding to a simple random walk (In particular, the $\theta_n$ are the shift operators on the underlying probability space $\Omega$. In trying to prove that $X_n-X_{n-1}$ are iid and uniformly distributed on $V=\{v\in \mathbb{Z}^d: |v|=1\}$, the lecturer claims the following equalities $$\begin{align}P_x[X_1-X_0=v_1,\ldots,X_n-X_{n-1}=v_n] &= \\ E^{P_x}[X_1-X_0=v_1,\ldots,\mathbb{1}_{\{X_1-X_{0}=v_n\}}\circ \theta_{n-1}] \\ \end{align} $$
I already have a problem with this, since to me the right hand side should be:
$$E^{P_x}[\mathbb{1}_{\{X_1-X_0=v_1\}}\cdots \mathbb{1}_{\{X_1-X_{0}=v_n\}}\circ \theta_{n-1}]$$ Then he computes $E^{P_x}[\mathbb{1}_{\{X_1-X_{0}=v_n\}}\circ \theta_{n-1}|\mathcal{F}_n]=\frac{1}{2d}$ and claims that $$ \begin{align} E^{P_x}[X_1-X_0=v_1,\ldots,\mathbb{1}_{\{X_1-X_{0}=v_n\}}\circ \theta_{n-1}] &= \\ E^{P_x}[X_1-X_0=v_1,\ldots,\mathbb{1}_{\{X_{n-1}-X_{n-2}=v_{n-1}\}}] \frac{1}{2d} \end{align} $$
I don't get this. How can he pass from expectation to conditional expectation just like that?