We know trace of matrix satisfies $$ \text{tr}(ABCD)= \text{tr}(DABC) $$ if the matrices are taking values on usually numbers.
Assume now the Matrices are Grassmannian valued, or they are fermions. By this, I mean a fermion taking values in color group such as SU(N). Then my question is: is it true that $$ \text{tr}(\psi_1\psi_2\psi_3\psi_4) = - \text{tr}(\psi_4\psi_1\psi_2\psi_3) $$ where the minus sign is due to anticommutate $\psi_4$ three times.
I prove it like this. \begin{eqnarray} \text{tr}(\psi_1\psi_2\psi_3\psi_4) &=& (\psi_1)^i_j (\psi_2)^j_k (\psi_3)^k_l (\psi_4)^l_i \\ &=& - (\psi_1)^i_j (\psi_2)^j_k (\psi_4)^l_i (\psi_3)^k_l \\ &=& (\psi_1)^i_j (\psi_4)^l_i (\psi_2)^j_k (\psi_3)^k_l \\ &=& - (\psi_4)^l_i (\psi_1)^i_j (\psi_2)^j_k (\psi_3)^k_l \\ &=& - \text{tr}(\psi_4\psi_1\psi_2\psi_3) \end{eqnarray}