Product of two epsilon tensors with n indices.

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I have spent a very long time trying to prove, or find a proof of, the formula for the product of two Lev-Civita/epsilon tensors:

$\epsilon _{i_1...i_n}\epsilon _{j_1...j_n}=n! \delta_{i_1j_1}...\delta_{i_nj_n}$

I can't seem to figure it out. And Wikipedia states the result but has not proven it (and I can't find this stated elsewhere).

I would like to know this proof because this result seems to be a necessary step in proving the determinant identity

$det(\textbf{A})\epsilon_{j_1...j_n}=\epsilon_{i_1...i_n}A_{i_1j_1}...A_{i_nj_n}$

And even if it is not necessary here, I would very much like to know!

EDIT: Apolgies for the confusion. The first equation I wrote I thought I had copied from Wikpedia: enter image description here

I must have misunderstood the notation because indeed, as was pointed out in a comment, what I have written would be either -1,0 or 1 on the LHS. I admit I just thought that superscript was the same as using subscript but was just there for clarity...

So returning to the determinant identity which I am trying to prove,

$det(\textbf{A})=\epsilon_{i_1...i_n}A_{i_11}...A_{i_nn}$

so

$det(\textbf{A})\epsilon_{j_1...j_n}=\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{i_11}...A_{i_nn}$

and I want to get this into the form

$det(\textbf{A})\epsilon_{j_1...j_n}=\epsilon_{i_1...i_n}A_{i_1j_1}...A_{i_nj_n}$

I had tried going about this in a couple of ways (forwards and backwards so to speak). The 'backwards' way started:

$det(\textbf{A})\epsilon_{j_1...j_n}=\epsilon_{i_1...i_n}A_{i_1j_1}...A_{i_nj_n}$

and so

$det(\textbf{A})\epsilon_{j_1...j_n}\epsilon_{j_1...j_n}=det(\textbf{A}) n!=\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{i_1j_1}...A_{i_nj_n}$

and then trying to show that the right side was

$\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{i_1j_1}...A_{i_nj_n}=n! \epsilon_{i_1...i_n}A_{i_11}...A_{i_nn}$

which would conclude the proof. But I don't know how to go about the last step, which is why when I saw the identity with two epsilon, and n! and Kronecker delta I thought that must be the key.

I have tried doing

$\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{i_1j_1}...A_{i_nj_n}=\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{i_11}...A_{i_nn}\delta_{1j_1}...\delta_{nj_n}$

which I'm not sure exactly what is wrong with it (I think it is because I am here picking values of the j_1...j_n rather than setting them to another variale, which is allowed. But in any case, this certainly leads to something very incorrect! Namely, $det(\textbf{A})n!=det(\textbf{A})$!