I'd write out \(\displaystyle (AB)^T_{i,j}\) and do a bit of rearranging of indices by noting for example \(\displaystyle A^T_{i,j}=A_{j,k}\)
and see where that takes you
\(\displaystyle (AB)_{i,j} = \sum \limits_k A_{i,k}B_{k,j}\)
\(\displaystyle (AB)^T_{i,j} = (AB)_{j,i} =
\sum \limits_k A_{j,k}B_{k,i} =
\sum \limits_k B_{k,i}A_{j,k}=
\sum \limits_k B^T_{i,k}A^T_{k,j} =
B^TA^T
\)