Lazard Lie ring of adjoint group of a radical ring equals associated Lie ring of the radical ring under suitable nilpotency assumptions

Adjoint group version
Suppose $$N$$ is a radical ring such that there exists a positive integer $$n$$ so that the additive group of $$N$$ has powering threshold at least $$n - 1$$ and $$x^n = 0$$ for all $$x \in N$$. Suppose $$N$$ has adjoint group $$G$$. Suppose that $$G$$ is a Lazard Lie group. Then, the Lazard Lie ring of $$G$$ is isomorphic to the Lie ring associated to $$N$$, i.e., the Lie ring whose additive group is the same as that of $$N$$ and whose Lie bracket is the additive commutator in $$N$$, i.e., $$[x,y] := xy - yx$$.

Algebra group version
Suppose $$q$$ is a power of a prime $$p$$ and $$N$$ is a nilpotent associative algebra over the finite field $$\mathbb{F}_q$$. Suppose, further, that $$x^p = 0$$ for all $$x \in N$$.

Let $$G$$ be the algebra group corresponding to $$N$$. Then, $$G$$ is a Lazard Lie group. Then, the Lazard Lie ring of $$G$$ is isomorphic to the Lie ring associated to $$N$$, i.e., the Lie ring whose additive group is the same as that of $$N$$ and whose Lie bracket is the additive commutator in $$N$$, i.e., $$[x,y] := xy - yx$$.