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We have that $\gamma_5 = -\frac{i}{4!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma$. Using this, what approach would be suggested in showing that $\gamma_5 \gamma^\sigma = \frac{1}{3!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho$? You would think that the solution is that \begin{equation} \gamma_5 \gamma^\sigma = -\frac{i}{4!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma \gamma^{\sigma} = -\frac{i}{4!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho \times 4 = \frac{1}{3!} \epsilon^{\mu \nu \rho \sigma} \gamma_\mu \gamma_\nu \gamma_\rho, \end{equation} which makes no sense...

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Indeed, you are right that conflating a saturated dummy index with a free index makes not sense.

However, you know that $$ \gamma_5 \gamma^\kappa= \tfrac{1}{2}[ \gamma_5 ,\gamma^\kappa ], $$ so what happens if you substitute the quadrilinear expression in the commutator? You know the leading, quintilinear, term vanishes. So only trilinears may survive. You can easily compute them. What do the 4 such add to?

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