Given the matrix
$$ A = \begin{bmatrix} 1&1&1&1\\ a&b&c&d\\ a^{2}&b^{2}&c^{2}&d^{2}\\ a^{3}&b^{3}&c^{3}&d^{3} \end{bmatrix} $$
How can I prove that $\det(A)=(d-c)(d-b)(d-a)(c-b)(c-a)(b-a)$ ?
And as a second question, what is the general formula for this type of matrix and how can I prove i?
I'm quite certain (though I could be wrong) that the general formula is as follows:
$$ \det( \begin{bmatrix} 1&\dots&1\\ a_{1}&\dots&a_{n}\\ \vdots&&\vdots\\ a_{1}^{n-1}&\dots&a_{n}^{n-1} \end{bmatrix} ) = \prod_{i>j}(a_{i}-a_{j}) $$ It definitely out for $n\leq4$.
I have tried a proof by induction over $n$. It's easy to prove that it holds for $n=1$:
$$ \det(\begin{bmatrix}1\end{bmatrix})=1\ \checkmark $$
For the inductive step, I've tried using row reduction on the last row and got to this:
$$ \det( \begin{bmatrix} 1&\dots&1\\ a_{1}&\dots&a_{n+1}\\ \vdots&&\vdots\\ a_{1}^{n}&\dots&a_{n+1}^{n} \end{bmatrix} ) = \sum_{k=1}^{n+1}(-1)^{n+k+1}\prod_{\begin{align}i&>j\\i,j&\neq k\end{align}}(a_{i}-a_{j}) $$
but I can't figure out how to prove that equals $\prod_{i>j}(a_{i}-a_{j})$. Is there a way to prove this. And if not, how should I go about solving this exercise?
If some of the mathematical notation I used is incorrect or very unusual, I would also appreciate it if that was pointed out to me.
