Jordan implies powers up to the fifth are well-defined

Statement
Suppose $$(S,*)$$ is a Jordan magma, i.e., it is a commutative magma satisfying the following identity:

$$\! (x * y) * (x * x) = x * (y * (x * x)) \ \forall \ x,y \in S$$

Then cubes, fourth powers, and fifth powers are well-defined in $$S$$. In other words, we have the following for all $$x \in S$$:

$$\! (x * x) * x = x * (x * x)$$

This quantity is denoted by $$x^3$$. Note that we will henceforth abbreviate $$x * x$$ as $$x^2$$.

$$\! x * x^3 = x^2 * x^2 = x^3 * x$$

This quantity is denoted by $$x^4$$.

$$\! x * x^4 = x^2 * x^3 = x^3 * x^2 = x^4 * x$$

This quantity is denoted by $$x^5$$.

Related facts for non-associative rings

 * Jordan implies power-associative: When we are dealing with non-associative rings instead of magmas, the condition of being a Jordan ring is strong enough to guarantee being a power-associative ring, which means that all powers (including sixth and higher powers) are well-defined.

Proof
Since $$S$$ is commutative, some of the equalities are direct. We do the proof in three stages.

The cube is well-defined
To prove: $$\! x^2 * x = x * x^2$$.

Proof: This follows from the fact that since the magma is commutative, $$x^2$$ and $$x$$ commute.

The fourth power is well-defined
To prove: <math! x * x^3 = x^2 * x^2 = x^3 * x.

Proof: Since the magma is commutative, we have $$x * x^3 = x^3 * x$$. It thus suffices to show the equality $$x * x^3 = x^2 * x^2$$.

To do this, we write:

$$\! x^2 * x^2 = (x * x) * (x * x) = x * (x * (x * x)) = x * x^3$$

In an intermediate step, we used Jordan's identity, setting $$y = x$$.

The fifth power is well-defined
To prove: $$x * x^4 = x^2 * x^3 = x^3 * x^2 = x^4 * x$$.

Proof: Since the magma is commutative, we obtain $$x * x^4 = x^4 * x$$ and $$x^2 * x^3 = x^3 * x^2$$. It thus suffices to show that $$x^3 * x^2 = x * x^4$$.

To do this, we write:

$$\! x^3 * x^2 = (x * x^2) * (x * x) = x * (x^2 * (x * x)) = x * x^4$$

In an intermediate step, we used Jordan's identity, setting $$y = x^2$$.