$x^{n}+1=0$ ------------$>(1)$
$\text{Case1:}$ If $n$ is odd $\text{(let n=3)}$
$x^{3}+1=0$
We can write like this
$x^{3}+0\cdot x^{2}+0\cdot x +1=0$ -----------$>(2)$
$(1-\alpha_{1})(1-\alpha_{2})(1-\alpha_{3})=1-\alpha_{1}-\alpha_{2}+\alpha_{1}\alpha_{2}-\alpha_{3}+\alpha_{1}\alpha_{3}-\alpha_{1}\alpha_{2}\alpha_{3}+\alpha_{2}\alpha_{3}$
$=1-\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{1}\alpha_{2}+\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{3}-\alpha_{1}\alpha_{2}\alpha_{3}$
$=1-(\alpha_{1}+\alpha_{2}+\alpha_{3})+(\alpha_{1}\alpha_{2}+\alpha_{2}\alpha_{3}+\alpha_{1}\alpha_{3})-\alpha_{1}\alpha_{2}\alpha_{3}$ -----$>(3)$
$\text{From equation $(2)$}$
$$\alpha_{1}+\alpha_{2}+\alpha_{3}=\frac{-b}{a}=\frac{0}{1}=0$$
$$\alpha_{1}\alpha_{2}+\alpha_{2}\alpha_{3}+\alpha_{1}\alpha_{3}=\frac{c}{a}=\frac{0}{1}=0$$
$$\alpha_{1}\alpha_{2}\alpha_{3}=\frac{-d}{a}=\frac{-1}{1}=-1$$
Put these value into equation $(3)$ we get
$(1-\alpha_{1})(1-\alpha_{2})(1-\alpha_{3})=1-(0)+(0)-(-1)=1+1=2$
$\text{Case2:}$ If $n$ is even $\text{(let n=2)}$
$x^{2}+1=0$
We can write like this
$x^{2}+0\cdot x +1=0$ -----------$>(4)$
$(1-\alpha_{1})(1-\alpha_{2})=1-\alpha_{1}-\alpha_{2}+\alpha_{1}\alpha_{2}$
$=1-(\alpha_{1}+\alpha_{2})+\alpha_{1}\alpha_{2}$ -------------$>(5)$
$\text{From equation $(4)$}$
$$\alpha_{1}+\alpha_{2}=\frac{-b}{a}=\frac{0}{1}=0$$
$$\alpha_{1}\alpha_{2}=\frac{c}{a}=\frac{1}{1}=1$$
Put these value into equation $(5)$ we get
$(1-\alpha_{1})(1-\alpha_{2})=1-(0)+1=2$
So$,$answer is $2$
Option $(D)$ is the correct choice$.$