Classic!

$\begin{aligned} (x-1)(y-1)(z-1) & = \color{#3D99F6}{xyz - (xy+yz+zx)} +(x+y+z) -1 \quad \quad \small \color{#3D99F6}{\text{From }\frac1x+\frac1y+\frac1z = 1 \implies xy+yz+zx = xyz} \\ & = \color{#3D99F6}{0} + \color{#D61F06}{x+y+z} - 1 \quad \quad \small \color{#D61F06}{\text{Applying AM-HM inequality}} \\ & \ge \color{#D61F06}{\frac9{\frac1x+\frac1y+\frac1z}} - 1 \quad \quad \small \color{#D61F06}{\text{Equality occurs when }x=y=z=3} \\ & \ge 9-1 = \boxed{8} \end{aligned}$

About AM-GM inequality .

Set $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$ , we'll have $a+b+c=1$ and the expression can be written as $\frac{(1-a)(1-b)(1-c)}{abc}=\frac{(b+c)(c+a)(a+b)}{abc}\stackrel{AM-GM}\geq\frac{8abc}{abc}=8$ The equality holds when $a=b=c=\frac{1}{3}$ or $x=y=z=3$