= Solution
The <Pauli matrix multiplication law> gives
$$
(\mathbf a\cdot\boldsymbol\sigma)(\mathbf b\cdot\boldsymbol\sigma)
=(\mathbf a\cdot\mathbf b)I+i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma.
$$
For a <unit vector> $\hat{\mathbf n}$, it follows that $(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I$. In the exponential series the even powers therefore multiply $I$ and the odd powers multiply $\hat{\mathbf n}\cdot\boldsymbol\sigma$. Summing them separately,
$$
\boxed{e^{-i\theta\hat{\mathbf n}\cdot\mathbf S}
=I\cos(\theta/2)-i\hat{\mathbf n}\cdot\boldsymbol\sigma\sin(\theta/2).}
$$
Equivalently, the <eigenvalues> of $\hat{\mathbf n}\cdot\boldsymbol\sigma$ are $\pm1$, so its two <spectral projectors> acquire phases $e^{\mp i\theta/2}$. The half-angle belongs to a <spinor> <rotation>, while the associated physical <spin> <vector> rotates through $\theta$.
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