An abstract group $G$ is called a $\mathbb Q$-powered group if for any $n \in {\mathbb N}$ and $g \in G$ there is an unique element $x \in G$ which is a solution of $x^n = g$. This "well" defines a right action $G \times {\mathbb Q} \rightarrow G$ (i.e. $g^r$ make sense for each $g \in G, r \in {\mathbb Q}$).
${\bf Usual ~ Hall-Petrescu :}$
$$ x^n y^n = (xy)^n c_2(x,y)^{n \choose 2} c_3(x, y)^{n \choose 3} \dotsc c_n^{n \choose n} $$ where $c_i(x,y)$ is some commutator in $x, y$ of weight $i$. Now for $r \in {\mathbb Q}$ define
$$ {r \choose k} := {\frac {r(r-1) \dotsc (r-k+1)}{k!}} $$
I need to prove that : if $G$ is a $\mathbb Q$-powered group with a $\mathbb Q$-powered normal subgroup $N$, then $G/N$ is also a $\mathbb Q$-powered group.
Apparently this need the general form of Hall-Petrescu formula for rational powers : which is replacing $n$ by a rational number $r$ in the usual Hall-Petrescu. In the book by Clement, Majewicz, Zyman (The theory of nilpotent groups, Theorem 4.16) they prove this for arbitrary $R$-powered groups where $R$ is a binomial ring (i.e., commutative integral domain with char$R = 0$ so that ${r \choose k} \in R$ for any $r \in R, k \in {\mathbb N}$) but using "Hall-Petrescu" (H-P) axiom (i.e. H-P holds for any $x, y \in G, r \in R$).
To apply this to $\mathbb Q$-powered groups we need to verify H-P axiom holds there. How do I prove this? A direct approach is too cumbersome.