Let $A$ be the Vandermonde matrix
$$ A=\begin{bmatrix} 1&1&\cdots&1 \\ a_1&a_2&\cdots&a_N\\ \vdots&\vdots&\ddots&\vdots\\ a_1^{M-2}&a_2^{M-2}&\cdots&a_N^{M-2}\\ a_1^{M-1}&a_2^{M-1}&\cdots&a_N^{M-1} \end{bmatrix} $$
Also assume that the matrix is fat i.e. $N>M$ and that all $a_i$ are distinct. Now, I attach a column to this matrix to get $B$ which is: $$ B=\begin{bmatrix} 1&1&\cdots&1&0 \\ a_1&a_2&\cdots&a_N&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_1^{M-2}&a_2^{M-2}&\cdots&a_N^{M-2}&0\\ a_1^{M-1}&a_2^{M-1}&\cdots&a_N^{M-1}&1 \end{bmatrix} $$ I want to prove that: $\mathrm{rank}(B)=M$. I know that $\mathrm{rank}(A)=M$ and that by attaching a column the rank will definitely not decrease so it may remain the same or increase. I am not comfortable with these types of arguments so what I am saying here might be insufficient for the proof. Any ideas?