A reference on the Baker-Campbell-Haudorff formula I'm in the process of writing my PhD thesis, and I'm in need of a reference I seem to be unable to find form the moment. I need a paper or a book treating the Baker-Campbell-Hausdorff formula which in particular proves that if $g$ is a dg Lie algebra, $x,y,z$ are three Maurer-Cartan elements, $\lambda$ is a gauge from $x$ to $y$, and $\mu$ is a gauge from $y$ to $z$, then $BCH(\lambda,\mu)$ is a gauge from $x$ to $z$.
I know the statement to be true, and have already seen it somewhere, but unfortunately I don't remember where. I tried to look it up in the works of Goldman-Millson and Getzler, but without success. I will be very grateful to anyone able to provide such a reference. Other references treating the BCH formula in general dg Lie algebras are also appreciated.
 A: We can prove the statement using the differential trick. We work over a field $\mathbb{K}$ of characteristic $0$ and with homological gradings.
Let $\mathfrak{g}$ be a dg Lie algebra. We write
$$\overline{\mathfrak{g}}:=\mathfrak{g}\oplus\mathbb{K}\delta\ ,$$
where $\delta$ has degree $-1$ with
$$[\delta,\delta] = 0\qquad\text{and}\qquad[\delta,x]=dx$$
for $x\in\mathfrak{g}$. Suppose $\alpha\in\mathrm{MC}(\mathfrak{g})$ is a Maurer--Cartan element, and let $x\in\mathfrak{g}$. The gauge action of $x$ on $\alpha$ is defined by the differential equation
$$\frac{d}{dt}\alpha(t) = dx + \underbrace{[\alpha(t),x]}_{\mathrm{ad}_x(\alpha)}$$
in $\mathfrak{g}$ with initial value $\alpha(0) = \alpha$. But this is equivalent to considering the differential equation
$$\frac{d}{dt}(\delta + \alpha(t)) = \mathrm{ad}_x(\delta + \alpha(t))$$
in $\overline{\mathfrak{g}}$. Its solution is then given by
$$\delta + \alpha(t) = e^{t\mathrm{ad}_x}(\delta + \alpha)\ ,$$
and thus the gauge action is
$$x\cdot(\delta + \alpha) = e^{\mathrm{ad}_x}(\delta + \alpha)\ .$$
Now if we have a second degree $0$ element $y$, then we obtain
$$x\cdot(y\cdot(\delta + \alpha)) = e^{\mathrm{ad}_x}e^{\mathrm{ad}_y}(\delta + \alpha) = e^{\mathrm{ad}_{\mathrm{BCH}(x,y)}}(\delta + \alpha) = \mathrm{BCH}(x,y)\cdot(\delta + \alpha)\ ,$$
as we wanted to prove.
