I am trying to get a complete proof of the following result (Ref. https://link.springer.com/article/10.1007/s00605-016-0938-5 Page-682, Lem.2.7):
${\mathrm{\bf Definition:}}$ Let $G$ be a group and $A$ be a $G$-module. A derivation $\delta : G \rightarrow A$ is a map that satisfies $\delta(gh) = \delta(g)^h \delta(h)$ for all $g, h \in G$ (written in multiplicative notation on $A$).
${\mathrm{\bf Lemma}}$: Let $F$ be a free group, $p$ be a prime number and $A$ be an $F$-module. Let $\delta : F \rightarrow A$ be a derivation. Then:
(i) $\delta(F^p) = \delta(F)^p [\delta(F),_{p-1} F]$,
(ii) if $A \leq F, [A,_{i} F] = 1$, we have $\delta(\gamma_i(F)) \leq [\delta(F),_{i-1} F]$ for all integers $i \geq 1$.
Proof. Let $x \in F$. Then $\delta(x^p) = \delta(x)^{x^{p-1}+x^{p-2}+ \dotsc + 1}$. Now $(x-1)^{p-1} \equiv x^{p-1} + x^{p-2} + \dotsc + 1$ mod $p$ in the group ring ${\mathbb{Z}}[F]$. So writing $x^{p-1} + x^{p-2} + \dotsc + 1 = (x-1)^{p-1} + u(x)$ we have $\delta(x^p) = \delta(x)^{u(x)} \delta(x)^{(x-1)^{p-1}}$. The element $\delta(x)^{u(x)} \in \delta(F)^p$ and $\delta(x)^{(x-1)^{p-1}} \in [\delta(F),_{p-1} F]$.
${\mathrm{\bf Question 1:}}$ Here the expression $\delta(F)^p$ must mean the subgroup (inside $A$) generated by the $p$-th power of the elements of the image $\delta(F)$. Otherwise, the proof does not work. So the right side of the equation (i) is a subgroup, whereas the left side is simply the image of $F^p$ (so, not necessarily a subgroup). How can these two be equal? Even if one presumes $\delta(F^p)$ is a subgroup, I don't see how the equality is proved.
Now, for (ii) assume it hold until $i=k$ and take $[A,_{k+1} F] = 1$. For $a \in F, b \in \gamma_k(F)$ we have
$$ \delta([a,b]) = [\delta(a), b] [a, \delta(b)]^{[a,b]} [a,b,\delta(a)] [a,b,\delta(b)] $$
Now by induction one can prove that for a subset $X \subseteq F$ we have $[X, \gamma_k(G)] \subseteq [X,_{k} F]$ and hence
$$ [a, \delta(b)]^{-1} [a, \delta(b)]^{[a,b]} = [[\delta(F), F], \gamma_{k+1}(F)] \subseteq [A,_{k+1} F] = 1 $$
So,
$$
\delta([a,b]) = [\delta(a), b] [a, \delta(b)] [a,b,\delta(a)] [a,b,\delta(b)]
$$
Now every term of the right side of above except $[a, \delta(b)]$ belong to $[\delta(F),_k F]$. I am stuck here.
${\mathrm{\bf Question 2:}}$ In the proof, the fact $F$ is a free group is not used. Shouldn't the proof work for any group $F$?