Here's one alternative method.
If $\ V_n\big(v_0,v_1,\dots,v_n\big)\ $ is the $n$-volume of the $n$-simplex $\ \mathcal{S}\ $with vertices $\ v_0,v_1,\dots,v_n\ $, then
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)\big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|}{n}\ ,
$$
where $\ P_n\ $ is the perpendicular projection onto the $\ n-1$-dimensional space spanned by $\ v_{n-1}-v_0,$$v_{n-2}-v_0,$$\dots, v_1-v_0\ $. The factor $\ \big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|\ $ in this product is just the height of the vertex $\ v_n\ $ above the $n-1$-simplex with vertices $\ v_0,v_1,\dots,v_{n-1}\ $, taken as a base of $\ \mathcal{S}\ $. By induction, therefore,
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{1}{n!}\|v_1-v_0\|\prod_{j=1}^{n-1}\big\|\big(I-P_j\big)\big(v_{j+1}-v_0\big)\big\|\ .
$$
Derivation of the formula
If $\ h_n=\big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|\ $ is the height of the vertex $\ v_n\ $ of the simplex $\ \mathcal{S}\ $ above its base, $\ \text{Conv}\big(v_0,v_1,\dots,v_{n-1}\big)\ $, and $\ A_{n-1}(x)\ $ is the $\ n-1$-volume of the cross-section of $\ \mathcal{S}\ $ at height $\ x\ $ above that base, then the volume of $\ \mathcal{S}\ $ is
$$
\int_0^{h_n}A_{n-1}(x)dx\ .
$$
But the cross-section of $\ S\ $ at height $\ x\ $ above the base is the $\ n-1$-simplex with vertices $\ v_0(x),v_1(x), \dots, v_{n-1}(x)\ $, where
$
v_i(x)=\left(1-\frac{x}{h_n}\right)v_i+\left(\frac{x}{h_n}\right)v_n
$. This simplex is similar to the base $\ \text{Conv}\big(v_0,v_1,\dots,v_{n-1}\big)\ $, scaled down by a linear factor $\ \frac{x}{h_n}\ $ $\big($since $\ v_i(x)-v_j(x)=\left(\frac{x}{h_n}\right)\big(v_i-v_j\big)\ $$\big)$. Its $n-1$-volume $\ A_{n-1}(x)\ $is therefore $\ \left(\frac{x}{h_n}\right)^{n-1}V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)\ $. The volume of $\ \mathcal{S}\ $ is therefore
\begin{align}
\int_0^{h_n}A_{n-1}(x)dx&=\int_0^{h_n}\left(\frac{x}{h_n}\right)^{n-1}V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)dx\\
&=\frac{h_n\,V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)}{n}\ ,
\end{align}
as stated above. The formula is a generalisation of the instances that the area of a triangle is half the length of its base multiplied by its height, and the volume of a tetrahedron is one third of the area of its base multiplied by its height.
Simplification
The above formula can be written more simply if the Gram-Schmidt procedure is used to orthonormalise the vectors $\ v_1-v_0,v_2-v_0,\dots, v_n-v_0\ $:
\begin{align}
u_1&=\frac{v_1-v_0}{\big\|v_1-v_0\big\|}\ ,\\
u_j&=\frac{v_j-v_0 -\sum_\limits{i=1}^{j-1}\left\langle v_j-v_0,u_i\right\rangle}{\big\|v_j-v_0 -\sum_\limits{i=1}^{j-1}\left\langle v_j-v_0,u_i\right\rangle\big\|}
\end{align}
for $\ j=2,3,\dots,n\ $. Then $\ \big(I-P_j\big)\big(v_{j+1}-v_0\big)=$$\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle u_{j+1}\ $, and $\ \big\|\big(I-P_j\big)\big(v_{j+1}-v_0\big)\big\|=$$\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle\ $. Therefore
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{\prod_\limits{j=0}^{n-1}\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle}{n!}\ .
$$
Postscript
It's occurred to me that the orthonormalisation, $\ u_1,u_2,\dots,u_n\ $, of $\ v_1-v_0,v_2-v_0,\dots, v_n-v_0\ $ allows one to define an isometry $\ v_0+\sum_\limits{i=1}^nx_iu_i\mapsto\big(x_1,x_2,\dots,x_n\big)\ $ from the affine hull of $\ \mathcal{S}\ $ onto $\ \mathbb{R}^n\ $. The image of $\ \mathcal{S}\ $ under this isometry will be an $n$-simplex in $\ \mathbb{R}^n\ $ congruent to $\ \mathcal{S}\ $, and therefore having the same volume. Therefore any of the well-known formulas for the volume of an $n$-simplex in $\ \mathbb{R}^n\ $ will give another method for calculating the volume of $\ \mathcal{S}\ $.
As $P$ is not a linear combination of $P_0$ and $P_\infty$, $\{P,P_0,P_\infty\}$ is a linearly independent set of degree $3$ polynomials. Let $B$ and $C$ be two points as above. We need to find a nonzero $Q=aP+bP_0+cP_\infty$ such that $Q$ vanishes at both $B$ and $C$. This is the same as solving the system
$$\begin{align}
\begin{bmatrix}
P(B)& P_0(B)&P_\infty(B)\\
P(C)& P_0(C)&P_\infty(C)
\end{bmatrix}
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
\end{bmatrix}
\end{align},$$
where $P(B)$ denotes $P$ evaluated at $B$ etc. A nontrivial solution $(a,b,c)$ exists in this case and hence a nonzero $Q$.
If $P$ were a linear combination of $P_0$ and $P_\infty$, we would only have a $2\times 2$ matrix, and there is no guarantee of a nontrivial solution to the system.
Best Answer
It's used when noting that $\Sigma$ is a $5 \times 2$ matrix (and hence has rank at most 2): what this matrix actually is is the vectors stacked together, i.e. if $A=(a_1,a_2)$ and so on, it's $$ \Sigma = \begin{pmatrix} 0 \\ A \\ B \\ C \\ D \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ a_1 & a_2 \\ b_1 & b_2 \\ c_1 & c_2 \\ d_1 & d_2 \end{pmatrix} . $$ Were we considering vectors in (say) 3 dimensions, the matrix $\Sigma$ would instead have rows $(a_1 , a_2, a_3)$ etc., so the rank cannot be bounded above by 2 any more and we can't do the $1+1+2 < 5$ part of the calculation that shows the rank is not $5$.