How to derive the Vandermonde Determinant? I watched this video https://www.youtube.com/watch?v=87iJTcXqTKY explaning the Vandermonde Determinant I understood everything but I was wondering why the guy never mentioned the (-1)^(i+j) term used in solving determinants?
 A: Step 1: Argue that the determinant of the Vandermonde matrix is a polynomial of degree $n-1$ in $x_1$.  This is argued by considering cofactor expansion.  If one were to actually compute the the determinant using cofactor expansion, there would be a ton of $(-1)^{i+j}$'s here.  However, instead of doing the computation, the video performs a "thought experiment" to determine the shape of the determinant.  And concludes that it is a polynomial of degree $n-1$ in $x_1$.
Interlude: The formula for cofactor expansion would be a mess to compute.  The goal of the video is figure out the determinant without going through the very long calculation of cofactor expansion.
Step 2: Determine that $x_1=x_2,\cdots,x_1=x_n$ are roots of the determinant.  This is done without actually computing the determinant, but, instead using the properties of the determinant.  If one had fully calculated the determinant in Step 1 and plugged in $x_1=x_2$, then the result would be zero, but since the video didn't actually compute the determinant, you must use other means to figure this out.  Since a polynomial of degree $n-1$ has $n-1$ roots (counted with multiplicities), this means that the determinant is of the form 
$$
g(x_2,\cdots,x_n)(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n).
$$
Here $g$ is some polynomial that does not depend on $x_1$.
Step 3: Figure out $g(x_2,\cdots,x_n)$.  Since $g(x_2,\cdots,x_n)$ is the coefficient of $x_1^{n-1}$, this can be calculated using cofactor expansion.  And can be seen to be another Vandermonde matrix.
