Exponential of nilpotent derivation with divided powers is automorphism

Statement
Suppose $$R$$ is a non-associative ring and $$d^{(i)}, i \in \mathbb{N}_0$$, is a nilpotent derivation with divided powers for $$R$$ with nilpotency $$n$$. Define the exponential of $$d$$ as the additive endomorphism of $$R$$ given by:

$$\exp(d) := d^{(0)} + d^{(1)} + d^{(2)} + \dots + d^{(n - 1)}$$

Then, $$\exp(d)$$ is an automorphism of $$R$$.

Note that $$d^{(0)}$$ is by definition the identity map.

Related facts

 * Exponential of derivation is automorphism under suitable nilpotency assumptions: If a derivation (in the ordinary sense, without divided powers) is nilpotent, and there are certain conditions on powering and torsion, then we can construct a nilpotent derivation with divided powers from it in a canonical fashion and take the exponential to get an automorphism.

Facts used

 * 1) uses::Exponential of nilpotent derivation with divided Leibniz condition powers is endomorphism: This shows that the exponential is an endomorphism. This part of the proof uses only the Leibniz rule-type conditions part of the definition of derivation with divided powers.
 * 2) uses::Exponential of nilpotent element with divided powers is invertible: This shows that the exponential is in fact invertible, hence an automorphism. This uses only the divided power part of the definition of derivation with divided powers.

Proof
Fact (1) shows that the exponential is an endomorphism. Fact (2) shows that it is invertible.