Exterior square of Lazard Lie group and Lazard Lie ring are in Lazard correspondence

Statement
Suppose $$G$$ is a Lazard Lie group and $$L$$ is its Lazard Lie ring. Then, the exterior square $$G \wedge G$$ is a Lazard Lie group, the exterior square $$L \wedge L$$ is a Lazard Lie ring and in fact, they are in Lazard correspondence with each other.

Further, the commutator map homomorphism $$G \wedge G \to [G,G]$$ is in Lazard correspondence with the Lie bracket homomorphism $$L \wedge L \to [L,L]$$.

Proof
The first thing we note before proceeding to the proof is that the argument is essentially a formal one in the following sense. The Lazard correspondence gives the formulas for group operations in terms of Lie ring operations and vice versa. We have formulas for the operations in the exterior square of both the group and the Lie ring.

$$\begin{array}{ccc} \mbox{Lazard Lie ring } L & \to & \mbox{Exterior square } L \wedge L\\ \downarrow & & \downarrow \\ \mbox{Lazard Lie group } G & \to &\mbox{Exterior square } G \wedge G \\ \end{array}$$

Specifically, our formulas allow for the translations:

$$\begin{array}{ccc} \mbox{Addition, Lie bracket in } L & \to & \mbox{Addition, Lie bracket in } L \wedge L\\ \downarrow & & \downarrow \\ \mbox{Group multiplication in } G & \to &\mbox{Group multiplication in } G \wedge G \\ \end{array}$$

The vertical arrows involve using the Baker-Campbell-Hausdorff formula for the Lazard correspondence. The horizontal arrows involve using the description of the exterior square in terms of the original Lie ring or group.

The goal is to show that the "diagram commutes" in the sense that both paths from $$L$$ to $$G \wedge G$$ give the same ultimate formulas.

We show this in two steps:


 * This is true for the Malcev correspondence over $$\mathbb{Q}$$, i.e., it is true if we ignore the issue of problematic denominators.
 * Since the ultimate formulas are independent of whether we are over $$\mathbb{Q}$$, it is always true.