z \frac{u^2\tau_z^2}{2}+\cdots$$ Then taking derivative with respect to $u$ we get $f'(u\tau_z)=z\circ(1+f(u \tau_z))\tau_z.$ Now for $z,\om_i\in\caS$ we have by Prop. \[prop:gdz\] and Prop. \[circ\] $$z*(\om_1\om_2\cdots \om_n)=\sum_{i=0}^n \om_1\cdots
\om_i\tau_{\om_i}(z)\tau_z(\om_{i+1}\cdots \om_n)
+\sum_{i=1}^n \om_1\cdots
 \om_{i-1}( z\circ \om_i)\tau_z(\om_{i+1}\cdots \om_n).$$ This yields $$\begin{aligned}
\ & z*\left(z^{\circ n_1}\frac{(u \tau_z)^{n_1}}{n_1!}\cdots
 z^{\circ n_d}\frac{(u \tau_z)^{n_d}}{n_d!}\right)\\
=& \sum_{i=0}^d \left(z^{\circ n_1}\frac{(u
\tau_z)^{n_1}}{n_1!}\cdots
 z^{\circ n_i}\frac{(u \tau_z)^{n_i}}{n_i!}\right)z\tau_z \left(z^{\circ n_{i+1}}\frac{(u
\tau_z)^{n_{i+1}}}{n_{i+1}!}\cdots
 z^{\circ n_d}\frac{(u \tau_z)^{n_d}}{n_d!}\right)\\
+&\sum_{i=1}^d
 \left(z^{\circ n_1}\frac{(u \tau_z)^{n_1}}{n_1!}\cdots
 z^{\circ (n_i+1)}\frac{(u \tau_z)^{n_i}}{n_i!}\right)\tau_z\left(z^{\circ n_{i+1}}\frac{(u
\tau_z)^{n_{i+1}}}{n_{i+1}!}\cdots
 z^{\circ n_d}\frac{(u \tau_z)^{n_d}}{n_d!}\right).\end{aligned}$$ Hence $$z *\big(1-f(u \tau_z)\big)^{-1}\tau_z(\bfone)=\frac{d}{du}\Big(\big(1-f(u
 \tau_z)\big)^{-1} \Big) (\bfone).$$ This implies that $$\exp_*(zu \tau_z ) (\bfone)=\big(1-f(u\tau_z)\big)^{-1} (\bfone)$$ as desired.

\[cor:key\] For all $z\in \caS$ we have $$\exp_*(\log_\circ (1+z\tau_z)) (\bfone)=\big(1-z\tau_z\big)^{-1} (\bfone).$$

If $z\in \zz$ then $\tau_z=\text{id}$ and therefore we have

\[cor:circ\] For $z\in \zz$ we have $$\exp(zu)=(2-\exp_{\circ}(z u) )^{-1},\quad \ \exp_*(\log_\circ
(1+z))=(1-z)^{-1}.$$

Let’s consider a non-trivial example of Cor. \[cor:key\]. Let $N=2$ and $z=y_{k,1}$ then we have $$\exp\left(\sum_{n=1}^\infty (-1)^{n-1}
\zeta\big(nk;(-1)^k\big)\frac{u^n}{n}\right)=1+\sum_{n=1}^\infty
 \zeta\big(\{k\}^n;\{-1\}^n\big) u^n$$ where $\zeta(s_1,\dots,s_d;(-1)^{\gs_1},\dots,(-1)^{\gs_d})=L_2(
s_1,\dots,s_d|\gs_1,\dots,\gs_d)$ are the alternating Euler sums. For example, by comparing the coefficients of $u^2$ and $u^3$ we get $2\zeta(\ol{k},\ol{k})=\zeta(\ol{k})^2-\zeta(2k),$ and $6\zeta(\ol{k},\ol{k},\ol{k})=\zeta(\ol{k})^3-3\zeta(\ol{k})\zeta(2k)+2\zeta(\ol{3k}).$ Here $\ol{s_j}$ means that the corresponding $\gs_j$ is odd.

The following two propositions are generalizations of Prop. 5-6 of [@IKZ] respectively whose computational proofs are mostly omitted since nothing new happens in our situation.

\[prop:zz’\] For 