Is this proof of the product of determinants in tensor notation correct? I'll start with the matrix C which is the product of the matrices A and B.
$$c^i_k = a^i_jb^j_k$$ The determinant of C is $$\frac{1}{3!}\delta_{ijk}^{rst} c^i_rc^j_sc^k_t $$
by the definition of multiplication plugging in and rearranging:
$$\frac{1}{3!}\delta_{ijk}^{rst} a^i_lb^l_ra^j_mb^m_sa^k_nb^n_t = \frac{1}{3!}\delta_{ijk}^{rst} a^i_la^j_ma^k_nb^l_rb^m_sb^n_t  $$
now the step I'm unsure of is that I know I can turn this into this on its own:
$$\delta_{ijk}^{rst} = \frac{1}{3!} \delta_{ijk}^{lmn} \delta_{lmn}^{rst} $$
But can I plug this into the equation and mix the dummy indices like this so I get: $$ \frac{1}{3!}\frac{1}{3!} \delta_{ijk}^{lmn} \delta_{lmn}^{rst} a^i_la^j_ma^k_nb^l_rb^m_sb^n_t $$ which should be the end of the proof since these are the determinants of A and B multiplied together.
 A: In your derivation you're using Einstein's notation (or Einstein summation convention), and it states that an index should not appear more than twice in any expression (if it appears once, then it's a free index, if twice, its a dummy index that you should sum it over all its values). 
exmple: $$a_{ki}b_{il}=\sum_\limits{i=1}^na_{ki}b_{il}=a_{k1}b_{1l}+a_{k2}b_{2l}+...+a_{kn}b_{nl} $$
here k and l are free indices and i is a dummy index.
You can prove that det(AB)=det(A).det(B) in this way :
$$det(A) = \frac{1}{3!} e^{lmn}e_{ijk} a_l^i a_m^j a_n^k $$
$$det(B) = \frac{1}{3!} e^{rst}e_{opq} b_r^o b_s^p b_t^q $$
and $$e_{lmn} e^{lmn} = 3!$$
$$\frac{1}{3!}e^{lmn}e_{lmn}det(A)=\frac{1}{3!} e^{lmn}e_{ijk} a_l^i a_m^j a_n^k$$
$$e_{lmn}det(A)= e_{ijk} a_l^i a_m^j a_n^k$$
starting from det(C) :
$$det(C)=det(AB)=\frac{1}{3!}e^{lmn}e_{ijk}(AB)_l^i (AB)_m^j (AB)_n^k=\frac{1}{3!}e^{lmn}e_{ijk}a_o^ib_l^oa_p^jb_m^pa_q^kb_n^q\\
=\frac{1}{3!}e^{lmn}(e_{ijk}a_o^ia_p^ja_q^k)b_l^ob_m^pb_n^q=\frac{1}{3!}e^{lmn}(e_{opq}det(A))b_l^ob_m^pb_n^q=det(A)det(B)$$
