# Proving multiplication with Dirac adjoint spinor is Lorentz scalar

I'm studying QFT with the book "Quantum Field Theory" by Dr. David Tong, and I have difficulties with understanding some used transformations.

On the page 88 the author calculates Hermitian conjugate of representation of Lorentz group using chiral representation of gamma matrices: $$S[\Lambda]^\dagger = \exp(\frac{1}{2}\Omega_{\rho \sigma}(S^{\rho \sigma})^\dagger)=\exp(-\frac{1}{2}\Omega_{\rho \sigma}\gamma^0S^{\rho \sigma}\gamma^0)=\gamma^0\exp(-\frac{1}{2}\Omega_{\rho \sigma}S^{\rho \sigma})\gamma^0=\gamma^0S[\Lambda]^{-1}\gamma^0$$ as far as $$(S^{\mu \nu})^\dagger = - \gamma^0S^{\mu \nu}\gamma^0$$,

where $$S^{\mu\nu} = \frac{1}{4}[\gamma^\mu,\gamma^\nu]$$.

I'm struggling with the equality $$\exp(-\frac{1}{2}\Omega_{\rho \sigma}\gamma^0S^{\rho \sigma}\gamma^0)=\gamma^0\exp(-\frac{1}{2}\Omega_{\rho \sigma}S^{\rho \sigma})\gamma^0$$, I have no idea how to perform such a transition

For a matrix, $$\exp A=1+A+\frac12 A^2+...$$ and the $$\Omega_{\rho\sigma}$$ are just scalar numbers, so it is nothing but $$\exp \gamma_0 S\gamma_0 = 1+\gamma_0 S \gamma_0 + \frac12\gamma_0 S \gamma_0 \gamma_0 S \gamma_0 +...$$ where I've omitted the suffix $$\sigma\rho$$ on $$S$$, then use $$\gamma_0^2=1$$.