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I need to prove that $\gamma_{5}^2=1$, and in order to do this I wrote:

\begin{equation} (\gamma_{5})^2=\gamma^{5}\gamma_{5}=\left(-\frac{i}{4!}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \right)\left(-\frac{i}{4!}\epsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} \right)=\frac{-1}{(4!)^2}\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}. \end{equation}

Using that $\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\rho\sigma}=4!$ and having already proved that $\gamma^{\nu}\gamma_{\nu}=4I_{4}$, $\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma^{\mu}=4\eta_{\nu\rho}I_{4}$ and $\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma}\gamma^{\mu}=-2\gamma_{\sigma}\gamma_{\rho}\gamma_{\nu}$, I arrived at:

\begin{equation} (\gamma_{5})^2=\frac{-1}{4!}\left(-2\gamma_{\sigma}\gamma_{\rho}\gamma_{\nu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\right)=\frac{2\cdot4^3}{4!}=\frac{16}{3}. \end{equation}

I have also tried the following:

\begin{equation} (\gamma_{5})^2=\gamma^{5}\gamma_{5}=\left(-\frac{i}{4!}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \right)\left(-\frac{i}{4!}\epsilon_{\alpha\beta\lambda\theta}\gamma^{\alpha}\gamma^{\beta}\gamma^{\lambda}\gamma^{\theta} \right)=\frac{-1}{(4!)^2}\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\lambda\theta}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \gamma^{\alpha}\gamma^{\beta}\gamma^{\lambda}\gamma^{\theta}=\frac{-1}{(4!)^2}\delta^{\mu\nu\rho\sigma}_{\alpha\beta\lambda\theta}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \gamma^{\alpha}\gamma^{\beta}\gamma^{\lambda}\gamma^{\theta}. \end{equation}

But I don't know how to proceed from this. What am I missing?

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    $\begingroup$ Hint: In the Einstein notation one is not supposed to repeat an index more than twice. $\endgroup$
    – Qmechanic
    May 1 at 11:29
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    $\begingroup$ Your indices are not consistent. You need to pick different names for the second product. $\alpha$, $\beta$, etc. $\endgroup$
    – joigus
    May 1 at 11:30
  • $\begingroup$ I tried to do so, but with so many indices I didn't know how to procede. $\endgroup$
    – Lemes
    May 1 at 11:31
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You can do this in a much simpler way: $$ (\gamma^5)^2 = (i \gamma^0 \gamma^1 \gamma^2 \gamma^3)^2 = - \gamma^0 \gamma^1 \gamma^2 \gamma^3\gamma^0 \gamma^1 \gamma^2 \gamma^3 $$ Now anti-commute the $\gamma$-matrices next to one another and use $\{\gamma^\mu, \gamma^\nu \}= 2 \eta^{\mu\nu}$.

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Perhaps, the simplest/fastest way:

In Dirac representation, $\gamma ^5$ as a product of the four gamma matrices is written as

$${\gamma ^5} = \left( {\begin{array}{*{20}{c}} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{array}} \right).$$

First check this familiar result: $\gamma^5=?= {\left( {{\gamma ^5}} \right)^\dagger }$. Now, try this $({\gamma ^5})^2=?$

About your approach and some hints for solving your problem in your last attempt:

The last part of your post is equivalent to

$${\left( {{\gamma ^5}} \right)^2} = -\frac{1}{{{{(4!\,)}^2}}}{\gamma ^a}{\gamma ^b}{\gamma ^c}{\gamma ^d}{\gamma _a}{\gamma _b}{\gamma _c}{\gamma _d}.$$

From ${\{\gamma ^a},{\gamma ^b}\} = 2{\eta ^{ab}}{I}$, you have to prove ${\{\gamma ^a},{\gamma _c}\} = 2\delta _c^a {I}$. Using this and also ${\gamma ^\mu}{\gamma_\mu}=4 I$, you can prove your final result.

Warning: You should avoid repeating a specific symbol more than two times since it collides with index symbols in the same term and strongly can cause confusion/mistake, as you performed in the middle of your post.

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