If $A^n=0$, then $\left(BAB^{-1}\right)^n=0$ If $A^n=0$, then $\left(BAB^{-1}\right)^n=0$. (Assume everything that's necessary for the products to make sense.) _Proof_ : $$\left(BAB^{-1}\right)^n=\underbrace{\left(BAB^{-1}\right)\left(BAB^{-1}\right)...\left(BAB^{-1}\right)}_{n-times}$$ $$=BA(B^{-1}B)A(B^{-1}B)...(B^{-1}B)AB^{-1}=BA(I)A(I)...(I)AB^{-1}$$ $$=BAA...AB^{-1}=BA^{n}B^{-1}=B(0)B^{-1}=0$$ This feels at about the same rigor as calculating a telescoping series... is there a better way that doesn't rely on the ellipses? Or is this okay?

In a slight variant on @Monadologie's answer, we can prove by induction on $n$, that $(BAB^{-1})^n=BA^nB^{-1}$, even without using $A^n=0$. (The base step $n=0$ is trivial, as is the inductive step$$(BAB^{-1})^k=BA^kB^{-1}\implies(BAB^{-1})^{k+1}=BA^kB^{-1}BAB^{-1}=BA^{k+1}B^{-1}.$$ _Then_ , using $A^n=0$, we're done.