Recall that
[tex]\det(\mathbf{AB})=\det\mathbf A\det\mathbf B[/tex]
[tex]\det\mathbf A=\det\mathbf A^\top[/tex]
for any square matrices [tex]\mathbf A,\mathbf B[/tex]. So if
[tex]\mathbf U^\top\mathbf U=\mathbf I[/tex], and [tex]\det\mathbf I=1[/tex], we have
[tex]\det(\mathbf U^\top\mathbf U)=(\det\mathbf U)^2=1\implies\det\mathbf U=\pm1[/tex]
