Proving that $\det{(A^i_j})= \sqrt{ |\det{(G)}|}$ Let $V$ be an $n$-dimensional vector space and let $(v_1, \dots, v_n)$ denote any oriented basis for $V$. Also, let $g$ be an inner product on $V$ and let $G$ denote the Gram matrix of inner products $G = [g(v_i, v_j)]$. I am trying to show that if $v_j = A^k_je_k$, where $e_k$ denotes a basis that is orthonormal with respect to g, then $\det{(A^i_j)} = \sqrt{G}$.
I believe I have found a useful intermediate result, but I'm not really sure how to close the deal. For vectors $v_i$ and $v_j$ we have:
$$
g(v_i, v_j) = g(A^k_i e_k, A^r_j e_r) = A^k_iA^r_j \delta_{kr} = \sum\limits_{m=1}^n A^m_iA^m_j = \langle A_i | A_j\rangle
$$
where $A_k$ denotes the $k^{th}$ column of $A$ and $\langle\cdot | \cdot\rangle$ denotes the standard Euclidean inner product. Therefore, the matrix $G$ is given by
$$
G = [\langle A_i|A_j \rangle]
$$
At this point, I'm not sure what to do. I'm thinking there's some essential fact I need to know in order to continue.
So, my question is, am I on the right track and if so what should my next step be?
Edit: I updated this question to change the assumption that the $e_i$ are the standard basis vectors to the assumption that the $e_i$ are orthonormal with respect to $g$
 A: In fact, you're almost done.
You have calculated that $G = A^{T}A$ (since you're working with the standard inner product!). Now $\det{G} = \det{(A^TA)} = \det{(A)}^2$ and taking square roots gives what you want.

The Gram determinant is uniquely characterized by the following properties.
Let $v_{k} : V^{k} = V \times \cdots \times V \to \mathbb{R}_{\geq 0 }$ be a map satisfying


*

*$v_k (a_{1},\ldots, a_{i-1}, \lambda a_{i}, a_{i+1}, \ldots,  a_{k}) = |\lambda| v_k(a_{1},\ldots, a_{k})$ for all $(a_{1},\ldots,a_k) \in V^{k}$, all $i$ and all $\lambda \in \mathbb{R}$.

*$v_{k}(a_{1}, \ldots, a_{i-1}, a_{i} + a_{j}, a_{i+1}, \ldots, a_k)$ for all $i \neq j$.

*$v_{k}(a_{1},\ldots,a_{k}) = 1$ if the $a_{i}$ are orthonormal.


Then $v_{k}(A) = \sqrt{\det{A^{T}A}}$, where $A$ is the $n \times k$ matrix with columns $A = (a_1,\ldots, a_k)$. Clearly, the expression $A \mapsto \sqrt{\det{A^TA}}$ has the desired properties.

Added: On 3Sphere's request, I'm sketching an argument.
First of all, note that a function satisfying the three properties above allows us to perform the following things while keeping track of the value of $v_{k}$:


*

*Multiply a column by a scalar.

*Adding one column to another.


These are the two things one needs to do to perform Gaussian elimination and Gram-Schmidt. If the vectors $a_1, \ldots, a_k$ are linearly dependent, then we can express one column as linear combination of the others using properties 1. and 2. of $v_k$, hence $v_k (a_1,\ldots, a_k) = 0$ (by property 1). Now Gram-Schmidt tells us that there is a way of transforming $(a_1,\ldots,a_k)$ into an orthonormal system spanning the same $k$-dimensional subspace, so $v_k (a_1, \ldots, a_k)$ is determined uniquely by property 3. 
Since the expression $v_k(A) = \sqrt{\det{(A^TA)}}$ has the three desired properties, we see that $v_k$ exists and is uniquely determined by these properties. 
Of course, $v_k(A)$ is nothing but the $k$-dimensional volume of the parallelepiped spanned by $(a_1,\ldots,a_k)$, and the proof is essentially the same as one of the proofs of the existence and uniqueness of the determinant.
The proof appears this way in the very nice German textbook Analysis 2 by K. Königsberger, but I don't think that there is a translation into other languages.
