Matrix exponentiation of Pauli matrix I was working through the operation of the time reversal operator on a spinor as was answered in this question, however, I cannot figure out how this step was done: 
$$e^{-i \large \frac{\pi}{2} \sigma_y} = -i\sigma_y.$$
I suspect it has something to do with a taylor series expansion. Here $\sigma_y$ is the pauli matrix which has the form $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$.

The relation is shown using a taylor series of the exponential:
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ so that $e^{-i\pi/2\sigma_y}$ can be expanded.
$e^{-\frac{i\pi\sigma_y}{2}}=1+\left(-\frac{i\pi\sigma_y}{2}\right)+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^2}{2!}+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^3}{3!}+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^4}{4!}+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^5}{5!}+...$
Noting that $\sigma_y^2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ then
\begin{equation}
\begin{aligned}
e^{-\frac{i\pi\sigma_y}{2}}&=1-i\sigma_y(\pi/2)-\frac{(\pi/2)^2}{2!}+i\sigma_y\frac{(\pi/2)^3}{3!}+\frac{(\pi/2)^4}{4!}-i\sigma_y\frac{(\pi/2)^5}{5!}+...\\
&=\bigg\{1-\frac{(\pi/2)^2}{2!}+\frac{(\pi/2)^4}{4!}+...\bigg\}-i\sigma_y\bigg\{(\pi/2)-\frac{(\pi/2)^3}{3!}+...\bigg\}\\
&=\cos(\pi/2)-i\sigma_y\sin(\pi/2)\\
&=-i\sigma_y
\end{aligned}
\end{equation}
Here the taylor series for cos and sin were used to simplify the infinite sequence:
$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...$ and
$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
