First Hitting Times of Brownian Motion have independent increments. As I read the proof of Theorem 7.4.1 in Rick Durrett: Probability: Theory and Examples:

I am confused about why a step is possible. The theorem with the proof is given in the linked picture. $T_a = \inf \{s \ge 0 \mid B_s = a\}$ is the first hitting time of a standard Brownian motion $B = (B_t)_{t\ge 0}$. $P_x$ is defined as
$$
P_x(B_{t_1} \in A_1,\dots B_{t_n} \in A_n) := P(B_{t_1}+x \in A_1,\dots B_{t_n}+x \in A_n),
$$
where $0 \le t_1 < \cdots < t_n$, $n\in \mathbb N$. $E_x(\cdot)$ denotes the corresponding expectation value. $\theta_s$ is the shift transformation. I can follow all the steps shown in the picture, without these one: Why can we pull out the conditional expectation in the last step of the last equation:
$$
E_0 \left(\prod_{i=1}^{n-1} F_i E_0(F_n \vert \mathcal F_{T_{a_{n-1}}}) \right) \stackrel{?}{=} E_0 \left(\prod_{i=1}^{n-1} F_i \right) E_0(F_n)
$$
Thank you for the answer :)
 A: Here is an explanation: From the intermediate step of the proof, we know that
$$ E_0(F_n\mid\mathcal{F}_{T_{a_{n-1}}}) = E_0(f_n(T_{a_n} - T_{a_{n-1}})\mid\mathcal{F}_{T_{a_{n-1}}}) = E_0(f_n(T_{a_n-a_{n-1}})) \qquad P_0\text{-a.s.} \tag{1} $$
Since the right-hand side of $\text{(1)}$ is constant, $E_0(F_n\mid\mathcal{F}_{T_{a_{n-1}}})$ is a constant random variable and hence is equal to its expectation:
$$ E_0(F_n\mid\mathcal{F}_{T_{a_{n-1}}}) = E_0(E_0(F_n\mid\mathcal{F}_{T_{a_{n-1}}})) = E_0(F_n) \qquad P_0\text{-a.s.} $$
Therefore it follows that
$$ E_0\left(\prod_{i=1}^{n} F_i\right)
= E_0\left(\prod_{i=1}^{n-1} F_i E_0(F_n \mid F_{T_{a_{n-1}}})\right)
= E_0\left(\prod_{i=1}^{n-1} F_i E_0(F_n)\right)
= E_0\left(\prod_{i=1}^{n-1} F_i\right) E_0(F_n). $$
A: The answer to my questenion is given in the comments:
In the first equal sign of the last equation, it is used that $F_n$ is $\mathcal F_{T_{a_{n-1}}}$ measurable. Then in the second step, the independence of $F_n$ and $\mathcal F_{T_{a_{n-1}}}$ is used.
