decreasing central series for Leibniz algebra In left Leibniz algebra we consider the decreasing series for Leibniz algebra $L$ as follows:
$$L^2=[L,L],\cdots , L^{n+1}=[L,L^n]$$
in [this][1] article author claims that:
$$[L^n,L]=[L,L^n]$$
     I suspect that the above claim is wrong, I could prove one side of this equation by Leibniz rule and showed that 
$[L^n,L]\subseteq  [L,L^n]$. How to show the other side of it? Does the other side have any counterexample or not?
[1]: "On some basic properties of Leibniz algebras", V.V. Gorbatsevich,  https://arxiv.org/abs/1302.3345
 A: It's usually called "descending" or "lower" central series (decreasing is probably a wrong Russian translation). You probably mean left Lebiniz algebra, i.e. satisfying the identity $ab.c=a.bc-b.ac$. This indeed clearly satisfies $L^2.L\subset L.L^2$. I think that, in contradiction with your claim this is a proper inclusion: in the free left Leibniz on $(a,b,c)$, $a.bc$ is in $L.L^2-L^2.L$. This can be seen in the finite-dimensional 3-nilpotent quotient $F=L/L^4$. 
The latter can be written as $F=F_1\oplus F_2\oplus F_3$, where $F_1$ has dimension 3 and basis $(a,b,c)$, $F_2$ has dimension 9 and basis $(aa,ab,ac,\dots,cb,cc)$, and $F_3=F.F^2$ has dimension 27 and basis all $x.yz$ when $(x,y,z)\in\{a,b,c\}^3$. This can been seen easily, since we can first consider the same in the free non-associative algebra $A$ (where $A_3$ then has dimension 54, with basis all $x.yz$ and $xy.z$ with $x,y,z$ basis elements, and then modding out by all Leibniz 27 relations between basis elements. Since these are linearly independent (all being of the form $xy.z=W$ for some $W\in A.A^2$), we get the given description.
Moreover, these relations come by pairs: $xy.z=x.yz-y.xz$, $yx.z=y.xz-x.yz$ (which are singletons for $x=y$, namely $xx.z=0$). Thus the dimension of $F^2.F$ is 9, with basis all $xy.z$ for $(x,y,z)\in\{a,b,c\}^3$ with $x<y$ in the ordering $a<b<c$. 
Actually, by the same argument already in the free left-Leibniz algebra $G$ on two generators $a,b$ this shows that $a.bb$ is not in $G^2.G$, and even in the free left-Leibniz on 1 generator $a$, the element $a.aa$ is not in $H^2.H$ ($\ast$).

($\ast$) essentially the same argument, but easier to check mechanically: consider the 3-dimensional algebra $L$ with basis $(x,y,z)$ and product $xx=y$, $xy=z$, and all other (seven) products being zero. Then it is left-Leibniz (all triple products of basis elements are zero, except $x.xx$, and the left-Leibniz identity for the triple $(x,x,x)$ yields $0=0$). Then $L^2$ is generated by $(y,z)$ and clearly $L^2.L=0$ while $L^3:=L.L^2$ is generated by $z$.
So indeed Gorbatsevich, p4 of his paper you link, seems to be wrong. In keeping with long-standing tradition of statements in Russian books with no proof or reference to proof, which sometimes turn out to be correct.
