Exponential of nilpotent element with divided powers

Inside an associative unital ring
Suppose $$R$$ is an associative ring that is also a unital ring, i.e., the multiplication in $$R$$ is associative and admits an identity element. Suppose $$\mathbf{x} = (x^{(1)},x^{(2)},\dots)$$ is a nilpotent element with divided powers. Set $$x^{(0)} = 1$$. The exponential of $$\mathbf{x}$$ is defined as:

$$\exp(\mathbf{x}) = \sum_{i=0}^\infty x^{(i)}$$

Note that in practice, the summation is finite because of the nilpotency assumption. Explicitly, if the nilpotency is $$n$$, then:

$$\exp(\mathbf{x}) = \sum_{i=0}^{n-1} x^{(i)}$$

In the case that $$R$$ satisfies some powering and torsion-freeness assumptions, $$x^{(i)} = x^i/i!$$, and in this case, taking $$x = x^{(1)}$$, we get:

$$\exp(\mathbf{x}) = \exp(x) = \sum_{i=0}^\infty \frac{x^i}{i!} = \sum_{i=0}^{n-1} \frac{x^i}{i!}$$

Between an associative ring and a related monoid
Suppose $$R$$ is an associative ring. Suppose $$\mathbf{x} = (x^{(1)},x^{(2)},\dots)$$ is a nilpotent element with divided powers. The exponential of $$\mathbf{x}$$ is defined as the following element of the formal set $$1 + R$$:

$$\exp(\mathbf{x}) = 1 + \sum_{i=i}^\infty x^{(i)}$$

Note that in practice, the summation is finite because of the nilpotency assumption. Explicitly, if the nilpotency is $$n$$, then:

$$\exp(\mathbf{x}) = 1 + \sum_{i=1}^{n-1} x^{(i)}$$

In the case that $$R$$ satisfies some powering and torsion-freeness assumptions, $$x^{(i)} = x^i/i!$$, and in this case, taking $$x = x^{(1)}$$, we get:

$$\exp(\mathbf{x}) = \exp(x) = 1 + \sum_{i=1}^\infty \frac{x^i}{i!} = 1 + \sum_{i=1}^{n-1} \frac{x^i}{i!}$$

Between a radical ring and its adjoint group
A special case of the preceding is where $$R$$ is a radical ring, so that $$1 + R$$ is the adjoint group of $$R$$. In this case, the exponential map is a map from nilpotent elements with divided powers in $$R$$ to the adjoint group of $$R$$.