Move each barred generator to the right and put \[q_i=\sum _{j=1}^nK_{ij}\psi _j\bar \psi _i, \qquad -\bar \psi K\psi =\sum _iq_i.\] Each \(q_i^2=0\), because every term repeats \(\bar \psi _i\). The \(q_i\) have even degree and commute, so finite expansion gives \[e^{\sum _iq_i}=\prod _{i=1}^n(1+q_i).\] Only \(q_1\cdots q_n\) can contribute to the top degree. A nonzero term in that product chooses every unbarred generator once, hence is indexed by a permutation \(\sigma \in S_n\).

The associated word is \(\psi _{\sigma (1)}\bar \psi _1\cdots \psi _{\sigma (n)}\bar \psi _n\). Moving all barred factors to the right gives \((-1)^{n(n-1)/2}\). Sorting the unbarred factors gives \(\operatorname {sgn}(\sigma )\). Returning the bars to the paired top order gives the same first sign again. Their product is \(\operatorname {sgn}(\sigma )\). Thus the required coefficient is \[\sum _{\sigma \in S_n}\operatorname {sgn}(\sigma ) \prod _{i=1}^nK_{i,\sigma (i)}=\det K.\]