chirality - Chiral symmetry of Dirac Equation

When studying the argument in the subject, I need to show that $\exp^{-i\alpha\gamma^5}\gamma^0=\gamma^0\exp^{+i\alpha\gamma^5}$. How can I do this? Should I write the Taylor expansion of the exponential and directly apply anticommutation? Is there a more direct way?

Answer

Another (equivalent) approach: for any matrix $A$, $$ \gamma^0 \mathrm e^{A}\gamma^0=\mathrm e^{\gamma^0 A\gamma^0} $$ as can be seen using $\mathrm e^{B^{-1}AB}=B^{-1}\mathrm e^A B$.

The property $\mathrm e^{B^{-1}AB}=B^{-1}\mathrm e^A B$ is one of the basic properties of the matrix exponential. There are multiple proofs of this property. For example, here there is a proof using the series expansion of the exponential. Maybe a more straightforward proof is $$ \mathrm e^{B^{-1}AB}=\lim_{n\to\infty}\left(1+\frac{B^{-1}AB}{n}\right)^n= \lim_{n\to\infty}B^{-1}\left(1+\frac{A}{n}\right)^nB=B^{-1}\mathrm e^AB $$