Volume of a $n$-simplex in $(n+k)$-dimensions From here the formula for calculating the volume of a $n$-simplex in an $n$-dimensional space is given. Please how does one find the volume of the same simplex existing in a $(n+k)$-dimensional space

$k>=1$.

For example, say we have the following vertices $$ v_{1}, v_{2}, v_{3}, v_{4}$$ from a 3-simplex, the volume would be
$$
 \frac{1}{3!}\begin{vmatrix}
1 & 1 & 1 & 1\\
v_{1} & v_{2} & v_{3} & v_{4}\\
\end{vmatrix} =
 \frac{1}{3!}\begin{vmatrix}
1 & 1 & 1 & 1\\
v_{1}.x & v_{2}.x & v_{3}.x & v_{4}.x\\
v_{1}.y & v_{2}.y & v_{3}.y & v_{4}.y\\
v_{1}.z & v_{2}.z & v_{3}.z & v_{4}.z\\
\end{vmatrix}
$$
Now my challenge is getting the volume of some $3$-simplex in any $(3+k)$-dimensional space e.g. say k=2, that would mean the volume of a 3-simplex existing in a 5-dimensional space

The $n$-volume of the $n$-simplex with vertices $\vec v_0, …, \vec v_n ∈ \mathbb R^{n + k}$ is
$$\frac{1}{n!}\sqrt{\det(\mathbf A^\mathsf T\mathbf A)}, \quad \text{where } \mathbf A = \begin{bmatrix}\vec v_1 - \vec v_0 & \vec v_2 - \vec v_0 & ⋯ & \vec v_n - \vec v_0\end{bmatrix},$$
because all the entries of $\mathbf A^\mathsf T \mathbf A$ are dot products $(\vec v_i - \vec v_0) \cdot (\vec v_j - \vec v_0)$ which are preserved under any isometry $\mathbb R^n → \mathbb R^{n + k}$, and this reduces to the usual formula $\frac{1}{n!} \lvert\det \mathbf A\rvert$ when $k = 0$.
(Note: the more symmetric formula using the matrix $\begin{bmatrix}
\vec v_0 & \vec v_1 & \vec v_2 & ⋯ & \vec v_n \\ 1 & 1 & 1 & ⋯ & 1
\end{bmatrix}$ only works in $n$ dimensions. I just fixed an incorrect claim otherwise on Wikipedia.)
