Proof that $\epsilon_{ijk}$ is the same as the determinant of the Kronecker delta I just learned about the Kronecker Delta function and $\epsilon_{ijk}$ for the first time, and I still can't wrap my mind around how to prove that $$\epsilon_{ijk} = det\begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix}$$ I understand that the RHS is  $$det\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=1$$ but how is that related to the fact that the $\epsilon$ tensor renders 1 for even, -1 for odd permutations of $i,j,k$ and $0$ for identical ones? And how can it be proven?
The only "proof" I could think of is$$ \delta_{i1}\delta_{j2}\delta_{k3} + \delta_{i2}\delta_{j3}\delta_{k1} + \delta_{i3}\delta_{j1}\delta_{k2} - \delta_{i3}\delta_{j2}\delta_{k1} - \delta_{i2}\delta_{j1}\delta_{k3} - \delta_{i1}\delta_{j3}\delta_{k2}  \\
    =  \begin{cases} 1 \cdot \delta_{ijk} & |\{ijk\} \in \{123\},\{312\},\{231\} \\ -1 \cdot \delta_{ijk} & |\{ijk\} \in \{213\},\{321\},\{132\} \end{cases} \\
    = \epsilon_{ijk} $$ but that doesn't look sound to me.
I am aware of this question, but it's kind of like two steps ahead when I'd like to be able to explain just the first.
Can somebody explain with an Einstein mindset?

A property of the determinant is:


*

*Exchanging two rows while leaving everything else unchanged changes the sign of the determinant, but not the magnitude.


This can be proven from the formula $\det(AB) = (\det A)(\det B)$. The operation of exchanging rows in a matrix $M$ is the same as taking $EM$, where $E$ is the same row exchange on the identity matrix $I$. One can easily show all such matrices $E$ have determinant $-1$.
Now note that if two rows of a matrix $M$ are identical, then exchanging them makes no change to the matrix. Therefore $\det M = -\det M$, from which it follows that $\det M = 0$:


*

*If a matrix has two identical rows, then its determinant is $0$.


Now define $$d_{ijk} = \det \begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix}$$
The matrix for $d_{123}$ is the identity matrix, so $d_{123} = 1$. If $i = j$ or $j = k$ or $i = k$, the matrix will have two identical rows, so $d_{ijk} = 0$ in this case. And by the exchange of row rule for matrix determinants, if you exchange the values of any two of $i, j, k$, then $d_{ijk}$ changes sign.
These three facts completely determine all values of $d_{ijk}$. But $\epsilon_{ijk}$ obeys exactly the same set of rules. So the two of them must be equal.
