Proving BCH formula 
I managed to solve a) by differentiation. But, I am stuck at b). I can't see why change from $B$ to $e^B$ puts the series of a) in the exponential function.  Also, I think I have to use a) and b) to show c). However I can't find a way to do so. Could anyone please explain to me?
 A: Ok here we go, using the identities given we clearly have
$$
\begin{align}
e^{sA}e^{sB} &= e^{sA} e^{sB} e^{-sA} e^{sA} \\
&= \exp\left(sB + s^{2}[A,B] + \frac{s^3}{2}[A,[A,B]] + \mathcal{O}(s^4) \right)e^{sA} \tag{1}
\end{align}
$$
and we want to show that after expansion this agrees, up to terms of $\mathcal{O}(s^4)$, with
$$
\exp\left(s(A+B) + \frac{s^2}{2}[A,B] + \frac{s^3}{12}\left([[A,[A,B]] + [B,[B,A]]\right) + \mathcal{O}(s^4)\right) \tag{2}
$$
the goal is just to expand both of these equations, collect terms and then show they agree. 
Worked example for the $s^2$ term
I will demonstrate for terms involving $s^2$ and let you try the others. So first define the argument of the exponential in $(2)$ by
$$
Z = s(A+B) + \frac{s^2}{2}[A,B] + \frac{s^3}{12}[[A,[A,B]] + [B,[B,A]]
$$
expanding the power series for the exponential and keeping only those terms involving $s^2$ we have
$$
I + Z + \frac{Z^2}{2} + \cdots \mapsto \frac{s^2}{2}[A,B] +\frac{s^2}{2}(A^2 + AB + BA + B^2) \tag{3}.
$$
So after defining
$$
X = sB + s^2[A,B] + \frac{s^3}{2}[A,[A,B]] + \mathcal{O}(s^4)
$$
the hope is that after similarly expanding $e^X e^{sA}$ and collecting terms the two polynomials agree. Now noting that we only need to consider terms up to $X^2$ consider
\begin{align}
\left( I + X + \frac{X^2}{2} \right)\left(I + sA + \frac{s^2A^2}{2} \right) \tag{4}
\end{align}
quick checks tell us that 
$$
X^2 = s^2B^2 +s^3B[A,B] +s^3[A,B]B+\mathcal{O}(s^4)
$$
so after carrying out the multiplication (4) and collecting only those terms involving $s^2$ we get
$$
\begin{align*}
\frac{s^2 A^2}{2} + s^2 B A + s^2[A,B] + \frac{s^2 B^2}{2} &= \frac{s^2}{2}[A,B] + \frac{s^2}{2} \left([A,B] + 2 BA + A^2 + B^2 \right) \\
&= \frac{s^2}{2}[A,B] + \frac{s^2}{2}\left(A^2 + AB + BA + B^2 \right)
\end{align*}
$$
which as hoped tells us that those terms with coefficients $s^2$ of both identities agree. 
I shall leave it to you to repeat the process for $s^0, s^1$ and $s^3$ applying the same ideas, hope that has helped. 
