Binomial formula for powers of a derivation

Statement
Suppose $$R$$ is a fact about::non-associative ring (i.e., a not necessarily associative ring -- this includes the associative ring, Lie ring, and other cases) and $$d$$ is a derivation of $$R$$. For any nonnegative integer $$n$$, if we denote by $$d^n$$ the composition of $$d$$ with itself $$n$$ times, with $$d^0$$ defined as the identity map.

We then have:

$$\! d^n(x * y) = \sum_{i=0}^n \binom{n}{i} \left(d^i(x) * d^{n-i}(y)\right)$$

Note that this in particular applies to the case of a fact about::derivation of a Lie ring and a fact about::derivation of an associative ring.

Applications

 * Exponential of derivation is automorphism under suitable nilpotency assumptions

Proof idea
The idea is to use the Leibniz rule (which establishes the case $$n = 1$$; $$n = 0$$ is vacuous) and prove by induction. The proof is almost exactly like the proof of the binomial formula for $$(x + y)^n$$ for a commutative associative algebra.

Proof details: case $$n = 0$$
In the case $$n = 0$$, both sides are $$x * y$$, so the equality holds vacuously.

Proof details: case $$n = 1$$
In this case, the left side is $$d(x * y)$$. The right side is:

$$\! \binom{1}{0} (d^0(x) * d^1(y)) + \binom{1}{1}(d^{1}(x) * d^{0}(y)) = (x * dy) + (dx * y)$$

By the Leibniz rule, the left side equals the right side.

Proof details: inductive step
Inductive hypothesis: $$\! d^{n-1}(x * y) = \sum_{i=0}^{n-1} \binom{n - 1}{i} d^i(x) * d^{n-1-i}(y)$$

To prove: $$\! d^{n}(x * y) = \sum_{i=0}^{n} \binom{n}{i} d^i(x) * d^{n-1-i}(y)$$

Proof: We have by definition:

$$\! d^n(x * y) = d\left(d^{n-1}(x * y)\right)$$

By the inductive hypothesis, we can expand the inside and we get:

$$\! d^n(x * y) = d\left\{\sum_{i=0}^{n-1} \binom{n - 1}{i} d^i(x) * d^{n-1-i}(y)\right\}$$

Since $$d$$ is additive, we can pull it inside the summation on the right side and get:

$$\! d^n(x * y) = \sum_{i=0}^{n-1}\binom{n - 1}{i}d\left\{d^i(x) * d^{n-1-i}(y)\right\}$$

Now using the Leibniz rule on the inside, we get:

$$\! d^n(x * y) = \sum_{i=0}^{n-1}\binom{n - 1}{i}\left\{(d^{i+1}(x) * d^{n-(i+1)}(y)) + (d^i(x) * d^{n-i}(y))\right\}$$

We now rearrange the summation and obtain:

$$\! d^n(x * y) = \sum_{i=0}^n \left(\binom{n - 1}{i} + \binom{n - 1}{i-1}\right)(d^i(x) * d^{n-i}(y))$$

We use Pascal's identity on the sum of binomial coefficients and obtain what we want:

$$\! d^{n}(x * y) = \sum_{i=0}^{n} \binom{n}{i} d^i(x) * d^{n-1-i}(y)$$