Powers of an element $x$ in a non-associative algebra

Question

Consider an element $x\in A$ where $A$ is an arbitrary algebra. Denote by $*$ its product. Consider $X^5$ as the set of containing all the different ways to write the product of $x$ by itself five times, for example $X^3$ would be $\{x*(x*x),(x*x)*x\}$.

Is it true that $X^6=(X^5*x)\cup(x*X^5)\cup\{(x*x)*((x*x)*(x*x))\}$ ?

My work: It seems more like a problem of group theory or combinatory but so far I have only proven that $X^4=(X^3*x)\cup(x*X^3)\cup\{(x*x)*(x*x)\}$ but as the power goes higher the possibilities gets out of hand. I'm hopefull that this is a more general result and appreciate any help.

edit: Here $x*X^5$ denotes the set of all $\alpha$ such that $\alpha=x*t$ where $t\in X^5$.

Answer 1

No. An example: $$ (x\ast (x\ast x))\ast ((x\ast x)\ast x) $$

Answer 2

You can count explicitly how many elements there are in $X^n$: this is the number of ways to put parenthesis in a $n$-letter word, which is the Catalan number $C_{n-1}$. See https://en.wikipedia.org/wiki/Catalan_number for more details.

In particular, $|X^6| = C_5 = 14$, while $|X^5| = C_4 = 5$.