Second-order term of Baker-Campbell-Hausdorff formula Let $G$ be a Lie group with Lie algebra $\mathfrak g$. I'm trying to prove the first two terms of the Baker-Campbell-Hausdorff formula for $\exp(tX)\exp(tY)$ satisfy $$
(\exp tX)(\exp tY) = \exp\big(t(X+Y) + \frac 1 2 t^2 [X,Y] + t^3 \hat Z(t)\big) 
$$
for some smooth function $\hat Z: (-\epsilon, \epsilon) \to \mathfrak g$ and some $\epsilon > 0$, for every $X, Y \in \mathfrak g$. 
What I've tried: Since $\exp : \mathfrak g \to G$ restricts to a diffeomorphism from a neighborhood of $0 \in \mathfrak{g}$ to a neighborhood of $e \in G$, the map $\phi : (-\epsilon, \epsilon) \to \mathfrak g$ defined by 
$$
\phi(t) = \exp^{-1}(\exp tX \exp tY)
$$
is smooth for some $\epsilon > 0$. So the problem amounts to proving $\phi''(0) = [X,Y]$ and applying Taylor's theorem. But I'm not sure how to do this without knowing the derivative of multiplication $G \times G \to G$ at general points $(g,h) \in G \times G$. I'm wondering if there's a more clever approach having something to do with the adjoint representations $\mathrm{Ad}: G \to \mathrm{GL}(\mathfrak g)$ and $\mathrm{Ad}_* = \mathrm{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak{g})$ since $\mathrm{ad}(X)Y = [X,Y]$. Any advice would be appreciated. 
 A: Well, the proof is a triviality (below) provided you've assumed Eichler's elegant proof that $\hat Z$ is in the Lie algebra, so, then, exponentials of t times a Lie algebra element are merely the first three terms in their Taylor expansion around t =0 ; all further lability can be stuck into $\hat Z$.
Then, trivially, defining $f(t)= e^{tX} e^{tY}$, so $f(0)=\mathbb{1}$, it follows that 
$$
\partial_t f(t)= (X+ e^{tX} Ye^{-tX})f(t)=(X+Y+t[X,Y]+ t^2 ~W(t)) f(t),
$$
for W the obvious smooth function in the Lie algebra, guaranteed by the Adjoint map $\operatorname{Ad}_{tX} Y$. 
As a representation of just the three terms of its Taylor expansion, this integrates to 
$$
f(t)= e^{t(X+Y) + t^2[X,Y]/2 + t^3 Z},
$$
for smooth undetermined functions W and Z. Eichler's proof guarrantees that Z is in the Lie algebra as well.

If you do not assume Z is in the Lie algebra, you might as well do the full enchilada (symbolically closed form) outlined in standard texts, such as W Miller's, or Varadarajan's, so, e.g., Application Algorithm 1 in my crib notes, or, even, WP.
Specifically, the exact answer for your desideratum right-hand-side is  the standard Poincaré  expression 
$$
e^{tX} e^{tY}=\exp \left (   tX +t\int_0^1 ds ~~\psi (\operatorname {Ad}_{\exp (tX)} \operatorname{Ad}_{\exp(stY)}) ~~Y     \right ),
$$ 
where $\psi(x)=\frac{x \ln x}{x-1}= 1- (1-x)/2 -1/(1-x)^3/6- ...$, related to the generating function of the Bernoulli numbers.
However, since $\operatorname{Ad}_{\exp(Z)}=\exp(\operatorname{ad}_Z)$ and you are not interested in terms cubic in t, so quadratic inside the integral, and $1-e^{\operatorname{ad}_{tX}}={\cal O}(t)$, one needs only keep $\psi(x)=1 -(1-x)/2 +...$ and ignore the ellipsis. This enormously simplifies the integrand.
The second term in the desideratum exponent then collapses to just 
$$
 t\int_0^1 ds ~~\psi (e^{\operatorname {ad}_{tX}} e^{ \operatorname{ad}_{stY}}) Y =t\int_0^1 ds ~~( 1+ \tfrac{1}{2} \operatorname{ad}_{tX} + \tfrac{1}{2}\operatorname{ad}_{stY}) Y + {\cal O}(t^2)\\ 
= t\int_0^1 ds ~~( Y +t[X,Y]/2) )+ {\cal O}(t^2)= tY +t^2[X,Y]/2+  {\cal O}(t^2), 
$$ 
as required in your desideratum. 
