Prove $\exp{i\frac{\pi}{2}(-1+\sigma_{i})}=\sigma_{i}$ 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.
$e^{\hat{X}+\hat{Y}}=e^{\hat{X}}.e^{\hat{Y}}.e^{-\frac{1}{2}[\hat{X},\hat{Y}]}$ 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,
$$
e^{i\frac{\pi}{2}(-1+\sigma_{i})}=(-i).(i.\sigma_{i})=\sigma_{i}
$$
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  $\ds{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}} = \sigma_{k}:\ ?}$


Note that 
$\ds{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}} =
-\ic\exp\pars{\ic\,{\pi \over 2}\,\sigma_{k}}}$. Lets consider
$\ds{\mrm{f}\pars{\mu} \equiv \exp\pars{\ic\mu\sigma_{k}}}$. Then,
\begin{align}
\mrm{f}'\pars{\mu} &amp; = \ic\sigma_{k}\,\mrm{f}\pars{\mu}\,,\qquad
\left\{\begin{array}{rcl}
\ds{\mrm{f}\pars{0}} &amp; \ds{=} &amp; \ds{\mathbf{1}\ \pars{~identity~}}
\\[2mm]
\ds{\mrm{f}'\pars{0}} &amp; \ds{=} &amp; \ds{\ic\sigma_{k}}
\end{array}\right.
\\[5mm]
\mrm{f}''\pars{\mu} &amp; = \ic\sigma_{k}\,\mrm{f}'\pars{\mu} = -\,\mrm{f}\pars{\mu}
\quad\implies\quad\mrm{f}''\pars{\mu} + \mrm{f}\pars{\mu} = 0
\end{align}

Then,
\begin{align}
\mrm{f}\pars{\mu} &amp; = A\sin\pars{\mu} + B\cos\pars{\mu}\quad\implies\quad
\left\{\begin{array}{rcrcl}
\ds{0\,A}&amp; \ds{+} &amp;\ds{B} &amp; \ds{=} &amp; \ds{\mathbf{1}}
\\[2mm]
\ds{A}&amp; \ds{+} &amp;\ds{0\,B} &amp; \ds{=} &amp; \ds{\ic\sigma_{k}}
\end{array}\right.
\\[5mm]
\implies\,\mrm{f}\pars{\mu} &amp; = \ic\sigma_{k}\sin\pars{\mu} + \mathbf{1}\cos\pars{\mu}
\implies
\color{#f00}{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}}} =
-\ic\,\mrm{f}\pars{\pi \over 2} = -\ic\pars{\ic\sigma_{k}} = \color{#f00}{\sigma_{k}}
\end{align}