Determinant of the matrix $\binom{m_i}{j-1}$ Let $m_1,\dots,m_n$ be real numbers $\ge n-1$.
How can I find the determinant of the matrix $A$ defined by $(a_{i,j})=\binom{m_i}{j-1}$, for $1\le i\le n$ and $1 \le j \le n$ ?

This all looks pretty VanDerMonde-ish, but I haven't found a good way to solve it.
 A: Result
$$\det \binom{m_i}{j-1} = \frac{1}{\prod_{k=1}^{n-1} k^{n-k}}\prod_{0<i_1<i_2\leq n}\left(m_{i_2}-m_{i_1}\right)$$

Proof
Let us simplify the proof by breaking it into several elementary facts.
Fact 1
$$\det \binom{m_i}{j-1} = Const_n\prod_{0<i_1<i_2\leq n}\left(m_{i_2}-m_{i_1}\right)$$
Proof 1
If we compute $\det \binom{m_i}{j-1}$ per definition, as a sum of the products of the elements taken with the appropriate sign, we will get a sum of products of type $m_1^{q_1}m_2^{q_2}\dots m_n^{q_n}$, where $\sum_{i=1}^n q_i \leq \frac{n(n-1)}{2}$ (each column will contribute $number\ of\ column-1$ to this sum). In other words, we will get a $n$-dimensional polynom, and the degree of each summand is at most $\frac{n(n-1)}{2}$.
On the other hand, if $\exists i_1,i_2 : m_{i_1}= m_{i_2}$, then 2 rows will be the same and $\det \binom{m_i}{j-1}$ is $0$. Thus, our determinant can be presented as 
$$\det\binom{m_i}{j-1} =SomePolynom\left(m_{1},\dots, m_n\right)\prod_{i_1< i_2}\left(m_{i_1}- m_{i_2}\right)$$
(it can be easily proved by absurd).
So we have two representations of the same polynom, the first one has "overall" degree of at most $\frac{n(n-1)}{2}$, the second one will have "overall" degree of at least $\frac{n(n-1)}{2}$. Thus, the "overall" degree of this polynom is exactly $\frac{n(n-1)}{2}$, and, equivalently, $$SomePolynom\left(m_{1},\dots, m_n\right)\equiv SomeConstant$$ This constant is the same for any given $n$, but can vary from $n$ to $n$, so we will add a subscript to recognise it: $Const_n$.
Fact 2
Recursive formula for $Const_n$ is $$Const_n = \frac{Const_{n-1}}{(n-1)!}$$
Proof 2
Let us compute the determinant of matrix $\binom{m_i}{j-1}$ by expanding it over the last column. 
We have
$$Const_n\prod_{0<i_1<i_2\leq n}\left(m_{i_2}-m_{i_1}\right)= \sum_{I=1}^n (-1)^{n+I}\binom{m_I}{(n-1)!} Const_{n-1}\prod_{0<i_1<i_2\leq n,\; i_1\neq I, i_2\neq I}\left(m_{i_2}-m_{i_1}\right)$$ 
We consider this as a polynom wrt $m_n$ and check the coefficient of $m_n^{n-1}$.
This coefficient from the left hand side expression is $Const_n \prod_{0<i_1<i_2< n} \left(m_{i_2}-m_{i_1}\right)$; the one taken from the right hand side is $(-1)^{n+n}\frac{1}{(n-1)!}Const_{n-1} \prod_{0<i_1<i_2< n} \left(m_{i_2}-m_{i_1}\right) = Const_{n-1} \prod_{0<i_1<i_2< n} \left(m_{i_2}-m_{i_1}\right)$. 
Since these values are equal, we have $$Const_{n} = \frac{Const_{n-1}}{(n-1)!}$$
Fact 3
$$Const_{2}=1$$
Fact 4
$$Const_{n} = \prod_{k=1}^{n-1} \frac{1}{k!} = \frac{1}{\prod_{k=1}^{n-1} k^{n-k}}$$
Proof 4
$\frac{Const_{n-1}}{Const_{n}} = \frac{\prod_{k=1}^{n-1} k^{n-k}}{\prod_{k=1}^{n-2} k^{n-k}}=(n-1)!$
