Chern-Simons computation I'm a little confused by the following computation from Baez and Muniain's Gauge Fields, Knots, & Gravity.

Let $M = [0,1]\times S$ where $S$ is any compact oriented manifold. Let $E \overset{\pi}{\to} M$ be a trivial vector bundle over $M$ with connection $A$ in ``temporal gauge", i.e if $t$ is the coordinate corresponding to $[0,1] \subseteq M$, then $A(\frac{\partial}{\partial t}) = 0$. We resolve the exterior derivative as $d = dt\wedge\partial_t + d_S$.
Why is
$$2\int_M\text{tr}(dt\wedge\partial_tA\wedge d_sA + dt\wedge\partial_t A\wedge A \wedge A) = \int_M\text{tr}\Big(dt\wedge\partial_t(A\wedge d_sA + \frac{2}{3}A\wedge A\wedge A)\Big)~?$$

It is only the first part of the RHS that I don't understand. i.e. I belive the the correspondence
$$2\int_M\text{tr}(dt\wedge\partial_tA\wedge d_sA) = \int_M\text{tr}\big(dt\wedge\partial_t(A\wedge d_sA)\big)$$
is off by a factor of two on the RHS.
 A: Let me use some shorthand notations (which are very popular among physicists)
$$d_t = dt \wedge \partial_t$$
Secondly I will not write the wedges. All the multiplications are to be understood as (noncommutative) wedge products.
In addition set the total exterior derivative over  $I\otimes M$
$$d = d_t + d_s$$
Since $d$, $d_t$, $d_s$, are exterior derivatives over manifolds, we have:
$$ d^2 = d_t ^2 = d_s^2 = 0$$
Therefore 
$$ d^2 = d_t^2 + d_t d_s + d_s d_t + d_s^2 = 0$$
Implies:
$$ d_s d_t + d_t d_s = 0$$
Also we need to remember the general cyclic trace identity ($B$ and $C$ are Lie algebra valued forms).
$$\mathrm{tr} (B \wedge C) = (-1)^{\mathrm{rank}(B)\mathrm{rank}(C)} \mathrm{tr} (C\wedge B)$$
The first term
$$ \begin{align*}
\mathrm{tr} (d_t  A d_s A) &=  \mathrm{tr} d_t ( A d_s A)- \mathrm{tr} (A  d_t  d_s A)\\
&=  \mathrm{tr} (d_t ( A d_s A))+ \mathrm{tr} (A  d_s d_t A) \\
&=  \mathrm{tr} d_t ( A d_s A)+\mathrm{tr}( d_s (A  d_t A)) -   \mathrm{tr} (d_s A  d_t A)\\
&=  \mathrm{tr} d_t ( A d_s A)+d_s\mathrm{tr}(A  d_t A) -   \mathrm{tr} (d_t A  d_s A)
\end{align*}
$$
Therefore:
$$2 \mathrm{tr} (d_t  A d_s A) =  \mathrm{tr} d_t ( A d_s A)+d_s\mathrm{tr}(A  d_t A)$$
The second term is an exact form (total derivative) over a compact manifold $M$ therefore it is identically zero.
Therefore:
$$2 \mathrm{tr} (d_t  A d_s A) =  \mathrm{tr} d_t ( A d_s A)$$
The second term
$$\mathrm{tr} (d_t ( A^3) )=  \mathrm{tr} ((d_t A)A^2 )- \mathrm{tr} (A(d_t A)A )+   \mathrm{tr} (A^2 (d_t A)) $$ 
Using the cyclic trace rule mentioned above, the second term has an additional $-1$ contribution, thus we have:
$$\mathrm{tr} (d_t ( A^3) )= 3 \mathrm{tr} ((d_t A)A^2$$ 
Remark: It is possible to use the trace rule because both $d_tA$ and $A^2 = A \wedge A$ are Lie algebra valued forms.
