I saw somewhere (I appreciate if anyone has any references to proof of this fact) that if $A$ is a finite dimensional associative algebra such that $\textrm{dim}(A)<n$, then $A$ satisfies the standard identity of degree $n$:
$$s\_n(x\_1,\dots, x\_n)=\sum\limits\_{\sigma\in S\_n}\textrm{sgn}(\sigma)x\_{\sigma(1)}\cdots x\_{\sigma(n)},$$
where $S\_n$ is the symmetric group of degree $n$. 


Is there a non-associative version of this result? I mean, if $A$ is a finite-dimensional (not necessarily associative) algebra, then $A$ satisfies some kind of "non-associative" standard polynomials? For example, maybe:
$$P\_1(x\_1)=x\_1=s\_1(x\_1),$$
$$P\_2(x\_1,x\_2)=x\_1x\_2-x\_2x\_1=s\_2(x\_1,x\_2),$$
$$P\_3(x\_1,x\_2,x\_3)=x\_1(x\_2x\_3)-(x\_1x\_2)x\_3-x\_1(x\_3x\_2)+(x\_1x\_3)x\_2-x\_2(x\_1x\_3)+(x\_2x\_1)x\_3+x\_2(x\_3x\_1)-(x\_2x\_3)x\_1+x\_3(x\_1x\_2)-(x\_3x\_1)x\_2-x\_3(x\_2x\_1)+(x\_3x\_2)x\_1=x\_1(x\_2x\_3-x\_3x\_2)+x\_2(x\_3x\_1-x\_1x\_3)+x\_3(x\_1x\_2-x\_2x\_1)+(x\_3x\_2-x\_2x\_3)x\_1+(x\_1x\_3-x\_3x\_1)x\_2+(x\_2x\_1-x\_1x\_2)x\_3$$
and etc... Thanks in advance for answers!



"Non-associative" standard polynomials
The first question, why the standard identity $s\_n$ holds in any $< n$-dim associative algebra is easy: This is a multilinear identity, so in order to check it for arbitrary $x\_1,...,x\_n$, it is enough to check it for basic elements $x\_1,...,x\_n$. Since the dim $<n$, two of the $x\_1,...,x\_n$ are equal, which makes $s\_n=0$. 


In the non-associative case the situation is slightly more complicated. For every elements $x\_1,....,x\_n$ of the algebra denote $x\_1\cdot ...\cdot x\_n=(...(x\_1x\_2)x\_3)...)x\_n$ (one can choose any other arrangement of parentheses. Then $s\_n$ is defined as usual, and the proof that $s\_n=0$ is an identity is as above.


