Convention: No sum over repeated indices unless otherwise stated.
Definition: DetA=e(ijk)a1ia2ja3k, sum, e(ijk) is Levi-Civita symbol.

A common proof* of detA=detAT, is incorrect. It is:

DetA=e(ijk)a1ia2ja3k=e(ijk)ai1aj2ak3=DetAT, sum

For Ex, the term
e(312)a13a21a32 becomes e(312)a31a12a23;
to show this is a term of D, you have to rearrange the as to get the first indices in numerical order which gives:
e(312)a12a23a31, which is a term of D if e(312) = e(231), which has to be proved in general (for each term).

Theorem:
Given e(ijk)ai1aj2ak3. Rearrange as so that first indices are in numerical order. Then:
e(ijk)ailaj2ak3=e(lmn)a1la2ma3n.

Proof (Mirsky):
Consider:
e(123)e(ijk)ai1aj2ak3
For every interchange of a pair of as, simultaneously change the corresponding indices of the first and second es, with the result that:
e(123)e(ijk)ai1aj2ak3 = e(lmn)e(123)a1la2ma3n, or
e(ijk)ai1aj2ak3 = e(lmn)a1la2ma3n
End Proof.

Therefore:
DetAT = e(ijk)ai1aj2ak3 = e(lmn)a1la2ma3n = DetA, sum

Note
Nothing changes for the general case ijk,

* Determinant of Transpose - ProofWiki