Calculus Integration Based Proof that the Volume of a Parallelepiped spanned by $\{v_1, \ldots, v_n\}$ is $\text{det}(v_1, \ldots, v_n)$? NOTE: This question is mostly setup to give context for the proof I am working on and how I want the proof to be. The only technical, and quite modest, question I have appears in the paragraph near the bottom of this post.
Let $v_1, \ldots, v_N \in \mathbb{R}^N$. A parallelepiped $P(v_1, \ldots, v_N) \subset \mathbb{R}^N$ is the set
$$
P(v_1, \ldots, v_N) = \left\{\sum_{i=1}^N t_iv_i \mid  0 \le t_i \le 1 \text{ for $i$ from 1 to $N$.}\right\}.
$$
It is well known that the volume of this set is given by
$$
|\text{det}(v_1, \ldots, v_N).|
$$
Let $1_P$ be the indicator function on the parallelepiped set defined above. I would like to define the volume of the parallelepiped as
$$
\text{vol}(P) = \text{vol}(P(v_1, \ldots v_N)) = \int_{\mathbb{R}^N}1_P dV.
$$
Let $\{e_1,\ldots, e_N\}$ be an orthonormal basis for $\mathbb{R}^N$. I would like to define the determinant as
$$
\text{det}(v_1, \ldots, v_N) = \sum_{\sigma \in S_N} \text{sgn}(\sigma) v_{1, \sigma(1)}\ldots v_{N,\sigma(N)}
$$
where $S_N$ is the symmetric group of size $N$ and $v_{i, j} = e_j \cdot v_i$ is the $j^{\text{th}}$ component of $v_i$.
I would like a proof that
$$
\text{vol}(P) = |\text{det}(v_1, \ldots, v_N)|
$$
using the definitions I've provided above for $\text{vol}$ and $\text{det}$.
I expect the proof to use the following facts.
(1) Let $Q = P(e_1, \ldots, e_N)$ denote the unit cube.  I'm comfortable to take it as proven that
$$
\text{vol}(Q) = |\text{det}(e_1, \ldots, e_N)| = 1
$$
(2) I'm also comfortable to take it as proven that for reasonable set $A \subset \mathbb{R}^N$ and linear transformation $T$ that $\text{vol}(TA) = \text{vol}(TQ)\text{vol}(A)$ because this can seemingly be proven from the appropriate manipulation of the partitions used in the definition of the Reimann or Darboux integral.
(3) I'm also comfortable to take it as proven there is an invertible transformation $T$ such that $P(v_1, \ldots, v_n) = T Q$
(4) I'm comfortable to take it as proven that any invertible transformation $T$ can be written as a product of elementary matrices $T = E_1\ldots E_k$.
(5) I'm comfortable to take it as proven that $\text{det}(T_1 \ldots T_k) = \text{det}(T_1)\ldots \text{det}(T_k)$.
I think all that remains to be proven is that $\text{vol}(EQ) = |\text{det}(E)|$ for all the elementary matrices. For the elementary matrices that either scale a single axis or swap two axes I am convinced of this.
All that remains to be proven is that the above statement holds for the elementary matrices that add a multiple of one axis to another. In parallelepiped language I am asking for a proof that shearing a cube along any axis does not change its volume.
Let $E^{\alpha \beta}(c)$ be the matrix with entries $E^{\alpha\beta}_{ij}(c)$ defined by
$$
E^{\alpha \beta}_{ij}(c) = \delta_{ij} + \delta_{i\alpha} \delta_{j\beta} c
$$
Can someone prove (or provide a link to a proof) that
$$
\text{vol}(E^{\alpha \beta}(c) Q) = 1
$$
Furthermore, I would expect this proof to involve an explicit calculus definite integration over coordinates. It is already intuitively clear to me that the "base" of parallelepiped has hyper-area of 1 and the "altitude" of the parallelepiped is 1, therefore the volume is also 1, but I would like to see a more rigorous proof of this fact. I guess the challenge for me here is getting the bounds on the definite integral right and carrying out the integration over the bounds.
Alternatively, if there is a proof which is as rigorous, but more elegant, taking a different approach, this would also suffice.
 A: This is just Cavalieri's principle in action, but I get it that you want to write down the integral. So, $$E^{\alpha\beta}(c)Q = \{x\in\Bbb R^N: 0\le x_i\le 1, i\ne\alpha,\  cx_\beta\le x_\alpha\le cx_\beta + 1\}.$$
By Fubini's Theorem, do the $\alpha\beta$ integral and then the rest over the usual cube in $\Bbb R^{N-2}$. Thus, the volume is
$$\int_0^1\int_{cx_\beta}^{cx_\beta+1}dx_\alpha\,dx_\beta = \int_0^1 x_\alpha\Big|_{cx_\beta}^{cx_\beta+1}\,dx_\beta = \int_0^1 1\,dx_\beta = 1,$$
of course, as required. (Drawing the two-dimensional parallelogram is, of course, helpful, but not necessary.)
A: We can actually make further reductions. Suppose $T:\Bbb{R}^n\to\Bbb{R}^n$ is the linear transformation
\begin{align}
T(x_1,\dots, x_n)&=(x_1+x_2,x_2,\cdots, x_n)
\end{align}
In terms of matrices, we're taking the second row of the identity matrix and adding it to the first row. We only need to restrict attention to this particular one, because we can perform row swaps to ensure we're only looking at rows 1 and 2, and then perform a scalar multiplication to ensure we only deal with $c=1$. Now,
\begin{align}
T(Q)&=\{\xi\in\Bbb{R}^n\,:\, \xi_2\leq \xi_1\leq \xi_2+1\,\quad\text{and}\quad \xi_2,\dots, \xi_n\in [0,1]\}
\end{align}
(i.e just put $\xi_1=x_1+x_2$, and $\xi_j=x_j$ for $j\geq 2$, and manipulate the inequalities $x_i\in [0,1]$ for all $i$ in terms of $\xi$). Now, we have
\begin{align}
\text{vol}(T(Q))&=\int_{T(Q)}1\,dV\\
&=\int_{[0,1]^{n-2}}\int_0^1\int_{\xi_2}^{\xi_2+1}1\,d\xi_1\,d\xi_2\, d(\xi_3,\dots, \xi_n)\tag{by Fubini}\\
&=\int_0^1(\xi_2+1-\xi_2)\,d\xi_2\\
&=1.
\end{align}
Here, it's clear that the integral over the last $n-2$ coordinates is trivially $1$ (this is just the $(n-2)$-dimensional volume of the cube $[0,1]^{n-2}$).
