Covariant gamma matrices

Indeed geometric interpretation of $\gamma_5$ is related to the volume form $$ V=\frac 1 {4!} \epsilon_{\mu\nu\alpha\beta} dx^\mu \wedge dx^\nu \wedge dx^\alpha \wedge dx^\beta = \frac 1 {4!} \sqrt{-g} \varepsilon_{\mu\nu\alpha\beta} dx^\mu \wedge dx^\nu \wedge dx^\alpha \wedge dx^\beta = \sqrt{-g} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 =\sqrt{-g} d^4x $$ Your $\gamma^5$ can be written as $$ \gamma^5 := \frac i {4!} \epsilon_{\mu\nu\alpha\beta} \;\gamma^\mu \gamma^\nu \gamma^\alpha \gamma^\beta \;, $$ which can be shown equivalents to \begin{eqnarray} \gamma^5 &=& i\;\sqrt{-\eta}\;\varepsilon_{0123}\; \gamma^0 \gamma^1\gamma^2 \gamma^3\;,\\ &=& i \gamma^0 \gamma^1\gamma^2 \gamma^3\;. \end{eqnarray} So, the most natural way to define $\gamma_5$ must be $$ \gamma_5 := \frac i {4!} \epsilon^{\mu\nu\alpha\beta} \;\gamma_\mu \gamma_\nu \gamma_\alpha \gamma_\beta \;, $$ Consequently, we have \begin{eqnarray} \gamma_5 &=& i \Big( \frac{-1}{\sqrt{-\eta}} \Big) \varepsilon^{0123} \;\gamma_0 \gamma_1 \gamma_2 \gamma_3 \;,\\ &=& -i \;\gamma_0 \gamma_1 \gamma_2 \gamma_3 \;,\\ &=& -i\; \gamma^0(-\gamma^1)(-\gamma^2)(-\gamma^3)\;,\\ &=& i \gamma^0 \gamma^1\gamma^2 \gamma^3\;\\ &=& \gamma^5 \end{eqnarray} So the position of 5 does not matter.