V is a variety of commutative semi group satisfying the identity $x^2 = x^3$. 
I need to prove that:
$|F\_V(\{x\_1\dots,x\_n\})|$ = $3^n -1$.
Any hints on this ?


$F\_V$ is V-free algebra.



Variety of commutative semi group
Let $S\_n = F\_V(\{x\_1, ..., x\_n\})$. Since $S\_n$ is commutative, its elements can be written in the form $x\_1^{r\_1} \cdots x\_n^{r\_n}$ where $r\_1 + \dots + r\_n > 0$. Morevover, since the semigroup $S\_n$ satisfies the identity $x^2 = x^3$, you may assume that each $r\_i$ is equal to $0$, $1$ or $2$. It follows that $|S\_n| \leqslant 3^n -1$. To prove that this inequality is in fact an equality, it suffices to verify that the set of size $3^n -1$
$$
\{x\_1^{r\_1} \cdots x\_n^{r\_n} \mid 0 \leqslant r\_i \leqslant 2 \text{ and } r\_1 + \dots + r\_n > 0\}
$$
equipped with the product defined by
$$
(x\_1^{r\_1} \cdots x\_n^{r\_n})(x\_1^{s\_1} \cdots x\_n^{s\_n}) = x\_1^{\min\{2, r\_1 + s\_1\}} \cdots x\_n^{\min\{2, r\_n + s\_n\}}
$$
is a semigroup of $V$.


