Exponential of nilpotent derivation with divided Leibniz condition powers is endomorphism

Statement
Suppose $$R$$ is a non-associative ring and $$d^{(i)}, i \in \mathbb{N}_0$$, is a nilpotent derivation with divided Leibniz condition 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 endomorphism of $$R$$.

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

Applications

 * Exponential of nilpotent derivation with divided powers is automorphism
 * Exponential of derivation is automorphism under suitable nilpotency assumptions
 * Truncated exponential of derivation is automorphism under suitable nilpotency assumptions

Proof
Given: $$d^{(i)}, i \in \mathbb{N}_0$$ is a nilpotent derivation with divided Leibniz condition powers for a non-associative ring $$R$$ and with nilpotency $$n$$. Define:

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

To prove: $$\sigma(x * y) = \sigma(x) * \sigma(y) \ \forall x,y \in R$$

Proof: It turns out that the proof is easier if we consider $$\sigma$$ as an infinite sum:

$$\sigma = d^{(0)} + d^{(1)} + d^{(2)} + \dots$$

while at the same time remembering that only finitely many summands are nonzero, hence the infinite sum makes sense.

With this convention, we consider the left side. The left side is:

$$\sigma(x * y) = \sum_{k=0}^\infty d^{(k)}(x * y)$$

Now, we use the Leibniz rule condition on each $$d^{(k)}$$ to obtain:

$$\sigma(x * y) = \sum_{k=0}^\infty \sum_{i + j = k; i,j \ge 0} d^{(i)}(x)d^{(j)}(y)$$

Let's now look at the original right side. This reads:

$$\sigma(x) * \sigma(y) = \sum_{i=0}^\infty d^{(i)}(x) * \sum_{j=0}^\infty d^{(j)}(y)$$

Now, we can see that the summations that we have reduced the left side and right side to are the same summation -- they both involve adding up all terms of the form $$d^{(i)}(x) * d^{(j)}(y)$$ for $$i,j \ge 0$$, although the additions occur in a different order. Note that there is in total only finitely many possible nonzero products we can get (at most $$n^2$$ since both $$i,j$$ must be less than $$n$$) so the two summations add up precisely the same finite list of terms. Thus, they are equal.