AoPS - Problem

I will use next lemma: If $A \in M_n(\mathbb{R})$ and $tr(A^k) = 0$ , $\forall k \in \{1,2, \ldots , n \}$ then $A^n = 0_n$. It can be solve with Newton sum. I try to prove that $tr((AB-BA)^k) = 0$ , $\forall k \in \{1,2, \ldots , n \}$ k = 1 is the property of trace. Now i consider $k \ge 2$. I have from the hypothsis that $x(AB - BA) = I_n - BA$ and $(x - 1) (AB - BA) = I_n - AB$. I have then $x^k (AB - BA)^k = (I_n - BA)^k$ and $(1 - x)^k (AB - BA)^k = (I_n - AB)^k$ Now we look at trace and we have $tr((I_n - BA)^k) = tr((I_n - AB)^k)$ and so we have $x^k tr( (AB - BA)^k) = (1-x)^k tr((AB - BA)^k)$. $x^k \neq (1-x)^k$ so we obtain $tr((AB - BA)^k)(x^k - (1-x)^k) = 0$ so $tr((AB - BA)^k) = 0$ for $k \ge 2$. So from lemma we have $(AB - BA)^n = 0_n$.