TITLE: Vandermond matrix generalisation for non-integer degrees
QUESTION [6 upvotes]: For a $n \times n$ Vandermonde matrix $$V:=\begin{bmatrix}1 & c_1 & c_1^2 & \cdots & c_1^{n-1} \\ 1 & c_2 & c_2^2 & \cdots & c_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & c_n & c_n^2 & \cdots & c_n^{n-1}\end{bmatrix}$$ we know that it is nonsingular if and only if $c_i \ne c_j$ for $i\ne j$.
I am curious if this property can be generalized for non-integer degrees. Suppose I am given a $n \times n$ matrix $$W:=\begin{bmatrix}c_1^{d_1} & c_1^{d_2} & c_1^{d_3} & \cdots & c_1^{d_n} \\ c_2^{d_1} & c_2^{d_2} & c_2^{d_3} & \cdots & c_2^{d_n} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_n^{d_1} & c_n^{d_2} & c_n^{d_3} & \cdots & c_n^{d_n}\end{bmatrix}.$$ Is it true that $W$ is nonsingular if and only if $c_i \ne c_j$ for $i\ne j$? If it matters, then I consider complex values for $c_i$ and real values for $d_i$.
If (as I guess) it is something well-known, could you, please, give me a reference?
Update: Obviously, I assume that $c_i \ne 0$ for all $i$. Let us also assume that $d_i \ne d_j$ for $i \ne j$.
Update 2: I have tried the following condition: for all $d_k\ne 0$ we have $c_i^{d_k} \ne c_j^{d_k}$ for $i \ne j$. It does not work. Actually, for $n=2$ the condition is $c_1^{d_2-d_1} \ne c_2^{d_2-d_1}$. This is satisfied, particularly, for $|c_1| \ne |c_2|$. 
Update 3: I have the following intuition. Let $\Delta_{ij}:=d_i-d_j$. The hypothesis: if for all $i\ne j$ we have $c_k^{\Delta_{ij}} \ne c_l^{\Delta_{ij}}$ for $k\ne l$, then $\det{W} \ne 0$. For integer $d$ we have exactly the Vandermond condition.

REPLY [3 votes]: I think I have a partial solution: suppose that all the $c_k$ are distincts and $>0$, and that the $d_k$ are distinct and real. Then the generalized determinant is not $0$. I leave to you the case $n=2$, then we proceed by induction.
The key point is the following: if we have an expression of the form
$$f(x)=\sum_{k=1}^n u_k\exp(v_kx)$$
with all the $v_k$ distincts, and if $f$ has $n$ distinct zeros $w_1<..<w_n$, then $f=0$ (and all the coefficients $u_k$ are zero). To see why, the case $n=1$ is trivial, and proceed by induction, multiplying $f$ by $\exp(-v_1x)$, and using that the derivative of $f$ has a zero in $]w_k, w_{k+1}[$ for $1\leq k\leq n-1$.
Now we return to the determinant; We write the terms on the last line in the form $\exp(d_k\log c_n)$, we replace $\log c_n$ by $x$, and developping with respect to the last line, we get an expression of the form $$W(x)=\sum_{k=1}^n b_k\exp(d_kx)$$
Note that $b_n$ is the determinant constructed with the $c_k$ and the $d_k$ with $k\leq n-1$. By the induction hypothesis, we have $b_n\not =0$.
Now obviously, $W(\log c_k)=0$ for $k=1,\cdots,n-1$. Suppose in addition that $W(\log c_n)=0$. This a contradiction with the property above, and we are done.