Prove $\exp{i\frac{\pi}{2}(-1+\sigma_{i})}=\sigma_{i}$

Question

How do we prove

$e^{{i\frac{\pi}{2}(-1+\sigma_{i})}}=\sigma_{i}$ ?

where $\sigma_{i}:$Pauli matrix and $1=$ Identity matrix

Note: I understand that $i\frac{\pi}{2}(-1+\sigma_{i})$ is anti-hermitian since $(-1+\sigma_{i})$ is hermitian, hence the exponential of it is Unitary.

Answer 1

$e^{\hat{X}+\hat{Y}}=e^{\hat{X}}.e^{\hat{Y}}.e^{-\frac{1}{2}[\hat{X},\hat{Y}]}$ if $[[\hat{X},\hat{Y}],\hat{X}]=0$ and $[[\hat{X},\hat{Y}],\hat{Y}]=0$.

Since $1$ and $\sigma_{i}$ commute, i.e, $i\frac{\pi}{2}[-1,\sigma_{i}]=0$, $$ e^{i\frac{\pi}{2}(-1+\sigma_{i})}=e^{-i\frac{\pi}{2}+i\frac{\pi}{2}\sigma_{i}}=e^{-i\frac{\pi}{2}}.e^{i\frac{\pi}{2}\sigma_{i}} $$

$$ e^{-i\frac{\pi}{2}}=cos({\pi}/{2})-isin(\pi/2)=-i $$

note that $\sigma_{i}^{2}=1$, hence

$$e^{i\frac{\pi}{2}\sigma_{i}}=cos(\pi/2)+i\sigma_{i}.sin(\pi/2)=i.\sigma_{i}$$

Substituting the terms, $$ e^{i\frac{\pi}{2}(-1+\sigma_{i})}=(-i).(i.\sigma_{i})=\sigma_{i} $$