Let $V$ be a finite dimensional vector space over a field of characteristic $0$, and let $T(V)$ be the tensor algebra (also called the free associative algebra) on $V$. This is actually a Hopf algebra, where the coproduct is defined on words by splitting them into two pieces in all possible ways: $$\Delta(v_{I})=\sum_{I=I'\amalg I''} v_{I'}\otimes v_{I''}.$$ Iterating the coproduct yields maps $\Delta^k:T(V)\to T(V)^{\otimes(k+1)}$ which break a word apart into $k+1$ pieces in all possible ways. Define the operator $\hat{\Delta}^k\colon T(V)\to T(V)^{\otimes (k+1)}$ which breaks a word apart into $k+1$ nontrivial pieces in all possible ways. Let $m^{k}\colon T(V)^{\otimes {k+1}}\to T(V)$ be the multiplication operator, and define $\zeta\colon T(V)\to T(V)$ by $$\zeta=\sum_{i=0}^\infty \frac{(-1)^i}{i+1}m^i\hat{\Delta}^i.$$ Because the operator $\hat\Delta^{i}$ is locally nilpotent, this will be a finite sum when applied to any given element of $T(V)$ so there is no need to worry about convergence issues. Also note that $m^0\hat{\Delta}^0$ is by convention the number of ways to split a word into one nontrivial piece, which means that it is the identity on $T^+(V)$ and trivial on the base field sitting in degree $0$. Writing everything out in terms of basic definitions, we have $$\zeta(v_I)=v_I-1/2\sum v_{I_1}v_{I_2}+1/3\sum v_{I_1}v_{I_2}v_{I_3}+\cdots, $$ where each sum is over all ways to split $I$ into nontrivial pieces $I_1,\ldots, I_k$. Low degree calculations show that
$\zeta(v)=v$
$\zeta(v_1v_2)=\frac{1}{2}[v_1,v_2]$
$\zeta(v_1v_2v_3)=\frac{1}{3}[v_1,[v_2,v_3]]-\frac{1}{6}[v_2,[v_1,v_3]]$
In particular, it looks like $\zeta$ lands in the space of iterated commutators, which in this case is the same as the primitive elements in the Hopf algebra. So this is my question:
How can one show that the image of $\zeta$ lies in the subspace of primitive elements?
Conceivably one could either show that $\Delta(\zeta(v_I))=1\otimes \zeta(v_I)+\zeta(v_I)\otimes 1$, or directly show that $\zeta(v_I)$ is an iterated commutator.