Suppose that p=0. Let x,y be two permutable elements of Gf. Then xy∈Gf and log(xy)=logx+logy.
The fact that xy∈Gf follows from Lemma 1. Let U be an open subgroup of the additive group L(G) and ϕ:U→ϕ(U) an exponential mapping of G with the properties of Proposition 3 of no. 2; U can be assumed to be small enough for log∣ψ(U)∣ to be the inverse mapping of ϕ. For n∈Z−{0} suitably chosen, xn∈ϕ(U), yn∈ϕ(U). Let u=logxn, v=logyn, whence xn=ϕ(u), yn=ϕ(v). By formula (2), [u,v]=0. The Hausdorff formula proves then that ϕ(λ(u+v))=ϕ(λu)ϕ(λv) for ∣λ∣ sufficiently small; hence, for every sufficiently large integer i,
ϕ(pi(u+v))=ϕ(piu)ϕ(piv)
that is
pi(logxn+logyn)=log(xnpiynpi)
or
npi(logx+logy)=npilog(xy).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.