Elegant Proof of the Product of Two Levi Cevita Tensors Is their an elegant way to prove the product of two Levi Cevita tensors is equivalent to a determinant of a matrix of Kronecker deltas? I know that the anti-symmetry and cyclic nature should be easily proved via determinant row-interchange laws, but is the proof inelegant when it comes to showing it zeroes out when two indices are repeated?
The property in mind is: $$\epsilon_{ijk}\epsilon_{lmn}=\det\left|
   \begin{array}{cccc}
      \delta_{il}   & \delta_{im}  & \delta_{in}   \\
      \delta_{jl}   & \delta_{jm}  & \delta_{jn}   \\
      \delta_{kl}   & \delta_{km}  & \delta_{kn}   
   \end{array}
\right|
$$

The determinante of $\left(A_{ij}\right)$ is defined by Leibniz as
$$\det \left(A_{ij}\right) := \sum_{\sigma}\text{sign}(\sigma)\Pi_i A_{i,\sigma(i)} $$
With permutations $\sigma$ and their sign $\text{sign}(\sigma)$ ($=+1$ if Permutation is an even of $1,2,3,4,...$, $-1$ if pertutation is odd, $0$ else)
This can be simplified with 
$$\det \left(A_{ij}\right)=\epsilon_{i_i,i_2,..}A_{1,i_1}A_{2,i_2}A_{3,i_3}\cdots $$
Introducing base vectors $e_i$ of a $n$ dimensional vector space  with $i=1,2,...,n$ and $e_ie_j=\delta_{ij}$ (Here $n=3$)
Than you can write your Matrix
$$\begin{pmatrix}      \delta_{il}   & \delta_{im}  & \delta_{in}   \\
      \delta_{jl}   & \delta_{jm}  & \delta_{jn}   \\
      \delta_{kl}   & \delta_{km}  & \delta_{kn}\end{pmatrix}=
\begin{pmatrix}e_i^\top\\e_j^\top\\e_k^\top\end{pmatrix}\begin{pmatrix}e_l&e_m&e_n\end{pmatrix}$$
The determinant of a matricies build of vectors is now
$$\det\begin{pmatrix}e_i^\top\\e_j^\top\\e_k^\top\end{pmatrix}=\epsilon_{ijk}\qquad \det\begin{pmatrix}e_l&e_m&e_n\end{pmatrix}=\epsilon_{lmn}$$
Using the product rule of determinats
$$\det\begin{pmatrix}      \delta_{il}   & \delta_{im}  & \delta_{in}   \\
      \delta_{jl}   & \delta_{jm}  & \delta_{jn}   \\
      \delta_{kl}   & \delta_{km}  & \delta_{kn}\end{pmatrix}=\det\begin{pmatrix}e_i^\top\\e_j^\top\\e_k^\top\end{pmatrix}
\cdot\det\begin{pmatrix}e_l&e_m&e_n\end{pmatrix}=
\epsilon_{ijk}\epsilon_{lmn} $$
PS. I am using the notation $\left(A_{ij}\right)$ is the matrix with bound indicies and $A_{ij}$ are their elements. 
PPS. This calculation can also be found on http://de.wikipedia.org/wiki/Levi-Civita-Symbol
