The answer is No. For $n=2$, there is the Hall identity $[[x,y]^2,z]=0$. Drensky proved in 1981 that these two identities (Hall and standard (=Amitsur-Levitzki)) form a basis of identities of $Mat_2(\Bbb C)$ (i.e. all identities are consequences of these two). For other $n$, the basis of identities is (probably) unknown.

Note that, according to the Kemer theorem, every associative algebra over a field of characteristic zero has a finite basis of identities.

The answer is No. For $n=2$, there is the Hall identity $[[x,y]^2,z]=0$. Drensky proved in 1981 that these two identities (Hall and standard (=Amitsur-Levitzki)) form a basis of identities of $Mat_2(\Bbb C)$ (i.e. all identities are consequences of these two). For higher $n$, bases of identities are (probably) unknown.

Note that, according to the Kemer theorem, every associative algebra over a field of characteristic zero has a finite basis of identities.

The answer is No. For $n=2$, there is the Hall identity $[x,y]^2,z]=0$. Drensky proved in 1981 that these two identities (Hall and standard (=Amitsur-Levitzki)) form a basis of identities of $Mat_2(\Bbb C)$ (i.e. all identities are consequences of these two). For other $n$ the basis of identities is (probably) unknown.

Note that according to the Kemer theorem every associative algebra over a field of characteristic zero has a finite basis of identities.

The answer is No. For $n=2$, there is the Hall identity $[[x,y]^2,z]=0$. Drensky proved in 1981 that these two identities (Hall and standard (=Amitsur-Levitzki)) form a basis of identities of $Mat_2(\Bbb C)$ (i.e. all identities are consequences of these two). For other $n$, the basis of identities is (probably) unknown.

Note that, according to the Kemer theorem, every associative algebra over a field of characteristic zero has a finite basis of identities.