Showing that a matrix of exponential form has determinant 1

Question

Let $\vec{\sigma}$ be Pauli matrices. Then it is said that $exp(i\vec{\theta} \cdot \vec{\sigma}/2)$ is a matrix belonging the $SU(2)$. Here $\vec{\theta}$ seems lke arbitary vectors in $\mathbb{R}^3$/ And However it is unclear to me that the determinant of $exp(i\vec{\theta} \cdot \vec{\sigma}/2)$ is indeed $1$. Could anyone explain why it is $1$? Also something related to this $\vec{\sigma}/2$ is called $SU(2)$ symmetry. What exactly is $SU(2)$ symmetry?

Answer 1

It is because, for every square matrix $M$, $\det e^M=e^{\operatorname{tr}M}$. Besides, the trace of the Pauli matrices is $0$.