Jordan algebra identities A Jordan algebra is a vector space with a commutative bilinear operation $\circ$ obeying an identity that's often written as 
$$   (x \circ y) \circ (x \circ x) = x \circ (y \circ (x \circ x)) . $$
I find this identity rather obscure.  If we write $x^2 = x \circ x$ and use $L_a$ to stand for left multiplication by $a$, we can rewrite it in a more appealing form:
$$  L_{x^2} L_x = L_x L_{x^2} .$$
However, I'd be even happier if this were a special case of a more general identity
$$  L_{x^m} L_{x^n} = L_{x^n} L_{x^m} \qquad (\ast) $$
holding for all $n, m \in \mathbb{N}$.   
This more general identity parses in any Jordan algebra, because any Jordan algebra is power-associative: expressions like $x \circ \cdots \circ x$ are independent of how you parenthesize them, so $x^n$ is well-defined.   But is this more general identity $(\ast)$ true in every Jordan algebra?
 A: This identity (*) is, indeed, true, and is, in fact, a step in one of the standard ways to prove that Jordan algebras are power-associative: see McCrimmon's 2004 book A Taste of Jordan Algebras, exercise 5.2.2A (question (2)) on page 201.
Edit: another reference, which has the better taste of not being an exercise and of being earlier in a book: Jacobson, Structure and Representations of Jordan Algebras (1968), page 35, just above formula (56).  (I also just realized that the fact is mentioned in the Wikipedia article on Jordan algebras, with that reference.)
For completeness of MathOverflow, I might as well copy the essence of the argument: first prove the identity
$$
L_{(b\circ d)\circ c} = L_{b\circ d}L_c + L_{c\circ d}L_b + L_{b\circ c}L_d - L_b L_c L_d - L_d L_c L_b
$$
by linearizing the Jordan identity (here we use the fact that the characteristic is not two).  Apply this to $b=a^k$ and $c=d=a$, giving
$$
L_{a^{k+2}} = 2L_{a^{k+1}}L_a + L_{a^2}L_{a^k} - L_{a^k} L_a^2 - L_a^2 L_{a^k}
$$
From this it follows by induction that all $L_{a^k}$ belong to the algebra generated by $L_a$ and $L_{a^2}$ and, since these commute, they all commute.
