# Row Reduction of matrix with distinct complex powers

I'm trying to reduce the following matrix given distinct complex numbers $$\beta_1, \beta_2,...,\beta_N$$ with the knowledge that $$N \geq M+1$$.

From my testing I can see that this reduces to an identity matrix with $$N=M+1$$, and if $$N > M +1$$ any rows after row $$M+1$$ become all zeroes.

But I'm unsure of how to fully show this in a convincing argument.

$$\begin{bmatrix} (\beta_1)^M& (\beta_1)^{M-1}& ... & \beta_1 & 1\\ (\beta_2)^M & (\beta_2)^{M-1} & ... & \beta_2 & 1 \\ \vdots &\vdots & &\vdots & 1 \\ (\beta_N)^M & (\beta_N)^{M-1} & ... & \beta_N & 1 \ \end{bmatrix}$$

Consider first the case where $$N = M + 1$$, and let $$A$$ be the given matrix. Further, let $$B_N$$ be the matrix with $$1$$s in the off-diagonal, and $$0$$s elsewhere, e.g. $$B_3 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}.$$ Then $$AB_N = \begin{bmatrix} 1 & \beta_1 & \cdots & \beta_1^M \\ 1 & \beta_2 & \cdots & \beta_2^M \\ \vdots & \vdots & \ddots & \vdots \\ 1 & \beta_N & \cdots & \beta_N^M \end{bmatrix}$$ is a Vandermonde Matrix, and is invertible if and only if $$\beta_1, \beta_2, \ldots, \beta_N$$ are all distinct. Since $$B_N$$ is its own inverse, it follows that $$A$$ is also invertible. Hence, row reduction will yield the identity matrix, as conjectured.
Now, if $$N \ge M + 1$$, then we can simply ignore rows $$M+2, M+3, \ldots, N$$, and row reduce the first $$M + 1$$ rows. By the above argument, they will also reduce down to the identity matrix. Now that a pivot exists in each column, one can easily reduce the rows below to $$0$$ rows.