= Solution
A complete pairwise <tensor contraction> of copies of $A$ is represented by closed index cycles. Each cycle is a <trace> $\operatorname{tr}(A^r)$, possibly with a sign from reversing an index pair. If a contraction is internal to one copy, it contains the already vanishing $\operatorname{tr}A$. Otherwise an odd total number of copies forces at least one cycle of odd length, since a sum of even lengths is even. That trace vanishes by the transposition argument in part (ii). Therefore \b[every such complete contraction is zero]. This proves <vanishing odd complete contractions of an antisymmetric matrix> without assuming that all contractions form a single trace.
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