Tr(AB) = Tr(BA)

Let matrix A have compnents A_ij, and B have components B_ij. Then AB has components A_ikB_kj, where singular occurances of an index are free and repeated ones are summed over (a convention which saves time). So:

Tr(AB)=A_ikB_ki=B_kiA_ik=Tr(BA)

And that's all there is to it (remember that Tr(C)=C_ii, the sum of the diagonal entries).
A very useful result for one so easy to prove, and important consequence being that conjugate matricies have the same trace

Tr(QAQ^-1)=Tr(QQ^-1A)=Tr(A)

so that the trace of a linear map is independent of its matrix representation.