Let $R$ be a ring with characteristic $0$. We can assume $R$ to be the $p$-adic field also, which is of course characteristic $0$. Let $f(x) \in R[[x]]$ be a power series and $f^{k}(x)$ be its $k$-th iteration.
Then we can divide $f^{k}(x)$ by$f^{(k-1)}(x)$ to get a power series \begin{equation} \phi(x)=\frac{f^{k}(x)}{f^{(k-1)}(x)} \ \cdots \cdots (1) \end{equation}
Now let us we define a $n$-dimesional power series with $n$-components $$g(X)=(g_1(X),~ g_2(X) , \cdots, g_n(X)) \in R[[X]]^{\color{red}{n}}, \ \text{where} \ X=(x_1,~x_2,\cdots, x_n).$$
Also consider the $k$-iteration of $g(X)$ by $g^k(X)=(h_1^k(X),~ h_2^k(X) , \cdots, h_n^k(X)),$ where $h_i^k(X)$ is the $i$-th component function obtained by iterating $g(X)$ through $k$-th times. Anyway this is not a big deal. But I am confusing myself whether we get similar division like equation $(1)$ above i.e., where $\exists$ a $\psi(X) \in R[[X]]^n$ such that $$ \psi(X)=\psi(x_1,~x_2, \cdots, x_n)=\frac{g^{k}(X)}{g^{(k-1)}(X)} \ \cdots \cdots (2)$$
Note 1: There exist ${\color{blue}{inverse}}$ power series of $g(X)$ and hence concept of inverse of $g^k(X)$ in $R[[X]]^{\color{red}{n}}$ also exists. In the same way, $g^{(k-1)}(X)$ has also inverse.
My question- Is equation $(2)$ is alright just like equation $(1)$ ?