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I want to figure out the trace of gamma matrices relating with $\gamma^{(d+1)}$ for even $d$ dimensional case.

First define $\gamma^{(d+1)}$ as \begin{align} \gamma^{(d+1)} = \gamma^1 \gamma^2 \cdots \gamma^d \end{align} What i want to obtain is \begin{align} tr[ \gamma^{(d+1)} \gamma^{\nu_1 \cdots \nu_n}] = \textrm{something} \end{align} by restricting $0 \leq n \leq d$, i think, there is some formula related with this.

What i know is \begin{align} tr[\gamma^{(d+1)}]=0 \end{align} I think there is some general formula related with \begin{align} \gamma^{(d+1)} \gamma^{\nu_1 \cdots \nu_n} =\cdots \end{align}

If you know the formula, please let me know. Actually i know for $d=4$, $tr[\gamma^5 \gamma^{\mu_1} \gamma^{\mu_2}\gamma^{\mu_3}\gamma^{\mu_4}] = -4i\epsilon^{\mu_1 \mu_2 \mu_3\mu_4}$ and of course the $6$ product($n>d=4$) of gamma matrices can be achieved recursively, but i am not sure about its generalization.

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Supergravity by Dan Freedman and Von Proeyen gives a procedure to expand the product of Gamma matrices with or without contraction in any dimension. An example is $\gamma^{\mu\nu\rho}\gamma_{\sigma\tau}=\gamma^{\mu\nu\rho}_{\sigma\tau}+6\gamma^{[\mu\nu}_{[\tau}\delta^{\rho]}_{\sigma]}+6\gamma^{[\mu}\delta^{\nu}_{[\tau}\delta^{\rho]}_{\sigma]}$ for any dimension. The numerical factors are basically the denominators that you get when you expand the antisymmetric brackets.

$\gamma^{d+1}$ is related to the highest ranked Clifford algebra element $\gamma^{\nu_1\nu_2...\nu_d}$ by the Levi-Civita symbol. Using this and the above method which you can find in section 3.1.4 of the above mentioned book, you can simplify the trace operation you have to trace of Clifford algebra elements(which is zero except for the highest ranked one) and a constant expression in terms of deltas which would give you the value of the trace.

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