If $G$ is a nilpotent group of class $2$, we know that $G' ≤ Z(G)$, and here we are taking the definition that class of nilpotency is the smallest length of the central series of $G$; where $G'$ is the derived group of $G$.
Prove that $[x^n,y]=[x,y]^n$
First I am trying for $n=2$
Now I have that $[x,y]^2=(xyx^{-1}y^{-1})(xyx^{-1}y^{-1})=xy(x^{-1}y^{-1}xy)x^{-1}y^{-1}=xyx^{-1}x^{-1}y^{-1}xyy^{-1}=xyx^{-1}x^{-1}y^{-1}x$ next what can we do? please write me in detail. Thanks..
Note that $G'\le Z(G)$ so $[x,y]$ commutes with every element of $G$. Hence $[x,y]^2=(xyx^{-1}y^{-1})(xyx^{-1}y^{-1})=((xyx^{-1}y^{-1})x)(yx^{-1}y^{-1}) = x(xyx^{-1}y^{-1})(yx^{-1}y^{-1})=x^2yx^{-2}y^{-1}=[x^2,y]$
We now use induction. Suppose $[x,y]^{n-1}=[x^{n-1},y]$. Then similarly \begin{eqnarray}[x,y]^n=[x,y] [x,y]^{n-1}= [x,y] [x^{n-1},y]=(xyx^{-1}y^{-1})(x^{n-1}yx^{-n+1}y^{-1})=((xyx^{-1}y^{-1})x^{n-1})(yx^{-n+1}y^{-1}) = x^{n-1}(xyx^{-1}y^{-1})(yx^{-n+1}y^{-1})=x^nyx^{-n}y^{-1}=[x^n,y]\end{eqnarray}