I was reading these lecture notes (NB: PDF):
For spin-1/2, the rotation operator $$ R_\alpha^{(s)}(\mathbf n)=\exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right) $$ can be written as an explicit $2\times2$ matrix. This is accomplished by expanding the exponential in a Taylor series: \begin{align} \exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right)&=1-\frac{i\alpha}{2}\left(\vec\sigma\cdot\hat{\mathbf n}\right)+\frac1{2!}\left(\frac{i\alpha}{2}\right)^2\left(\vec\sigma\cdot\hat{\mathbf n}\right)^2\\&\quad-\frac1{3!}\left(\frac{i\alpha}{2}\right)^3\left(\vec\sigma\cdot\hat{\mathbf n}\right)^3+\cdots \end{align} Note that $$ \left(\vec\sigma\cdot\hat{\mathbf n}\right)^2=\left(\vec\sigma\cdot\hat{\mathbf n}\right)\left(\vec\sigma\cdot\hat{\mathbf n}\right)=\hat{\mathbf n}\cdot\hat{\mathbf n}+i\sigma\left(\hat{\mathbf n}\times\hat{\mathbf n}\right)=1 $$ Thus, the Taylor series becomes \begin{align} \exp\left(-i\frac{\alpha}{2}\vec\sigma\cdot\mathbf{\hat n}\right)&=1-\frac{i\alpha}{2}\left(\vec\sigma\cdot\hat{\mathbf n}\right)+\frac1{2!}\left(\frac{i\alpha}{2}\right)^2\left(\vec\sigma\cdot\hat{\mathbf n}\right)^2\\&\quad-\frac1{3!}\left(\frac{i\alpha}{2}\right)^3\left(\vec\sigma\cdot\hat{\mathbf n}\right)^3+\cdots\\ &=\left[1-\frac1{2!}\left(\frac\alpha2\right)^2+\frac1{4!}\left(\frac\alpha2\right)^$+\cdots\right]\\&\quad-i\vec\sigma\cdot\hat{\mathbf n}\left[\left(\frac\alpha2\right)-\frac1{3!}\left(\frac\alpha2\right)^3+\cdots\right]\\ &=\cos\left(\frac\alpha2\right)-i\vec\sigma\cdot\hat{\mathbf n}\sin\left(\frac\alpha2\right) \end{align}
However, the part I don't understand is:
$$ \left(\vec\sigma\cdot\hat{\mathbf n}\right)^2=\left(\vec\sigma\cdot\hat{\mathbf n}\right)\left(\vec\sigma\cdot\hat{\mathbf n}\right)=\hat{\mathbf n}\cdot\hat{\mathbf n}+i\sigma\left(\hat{\mathbf n}\times\hat{\mathbf n}\right)=1 $$
Why is that equal to 1? Where do the dot-product and cross-product come from? Note that the $\sigma$ are Pauli spin matrices.