Conditional expectation: when does $X_t=E[X_t\mid \mathcal{F}_s]$ for $sI came across a calculation (1$^\circ$ calculation, 2$^{nd}$ step) that stated, for $s<t$
$$E[B_s(B_t^2-t)]=E[B_sE[(B_t^2-t)\mid\mathcal{F}_s]]$$
I know the expectation here is zero, however, I don't get how $B_t^2-t$ is replaced by $E[(B_t^2-t)\mid\mathcal{F}_s]$. I feel like this is a basic property of probability that I should know. Or is it unique to this particular martingale ($B_t^2-t$) or just a property of Brownian motion? Or is it only because it is inside another expectation?
In general, for $s<t$, when does $X_t=E[X_t\mid \mathcal{F}_s]$? I know this is the case when $X_t$ is $\mathcal{F}_s$-measurable, but this doesn't seem to be the case here since $s<t$. I know that $X_s=E[X_t\mid \mathcal{F}_s]$ when $s<t$ and $X$ is a martingale, of course.
Maybe a step-by-step would be: 
$$
\begin{aligned}
E[B_s(B_t^2-t)]&=E[B_sE[(B_t^2-t)\mid \mathcal{F}_t]] \\
&=E[B_sE[(B_t^2-t)\mid \mathcal{F}_t]\mid\mathcal{F}_s] \quad \text{(why?)}\\
&=E[B_sE[(B_t^2-t)\mid \mathcal{F}_s]\mid\mathcal{F}_t] \quad \text{(tower property?)}\\
&=E[B_sE[(B_t^2-t)\mid\mathcal{F}_s]]
\end{aligned}
$$
I get the last step, but don't know why or if the first three steps would be valid.

Let's work it out (also let's finish the calculation):
$$
\begin{aligned}
E[B_s(B_t^2-t)]&=E[E[B_s (B_t^2-t) \mid \mathcal{F}_s]] \\
&=E[B_s E[(B_t^2-t) \mid \mathcal{F}_s]] \\
&=E[B_s (B_s^2-s)] \\
&=E[B_s^3]-E[sB_s] \\
&=0.
\end{aligned}
$$
The first equality is the tower property. The second equality is "factoring out what is known". The third equality is the martingale property for $B_t^2-t$. The rest is linearity and properties of Gaussian random variables.
