About the determinant of a $4\times 4$ Vandermonde matrix I'm struggling with proving the Vandermonde matrix of dimension 4x4.
I don't want to get into induction, if that is possible.
I know there is a lot of material on the internet but I am looking for a calculation solution, and not an induction one.
I have reached this expression: 
$$
a_1^3(a_4-a_3)(a_3-a_2)(a_4-a_2)-a_2^3(a_4-a_3)(a_4-a_1)(a_3-a_1)+a_3^3(a_4-a_2)(a_4-a_1)(a_2-a_1)-a_4^3(a_3-a_2)(a_3-a_1)(a_2-a_1)
$$
Obviously this is a very long expression to simplify.
Thanks in advance.
 A: It's easier to prove the general formula, with some simple results on polynomials in several indeterminates. Consiider the Vandermonde determinant:
$$\begin{vmatrix}
1&1&\dots&1\\
a_1&a_2&\dots&a_n\\
a_1^2&a_2^2&\dots&a_n^2\\
\vdots&&&\vdots\\
a_1^{n-1}&a_2^{n-1}&\dots&a_n^{n-1}
\end{vmatrix}$$
The general formula for the determinant: 
$$\sum_{\sigma\in S_n}(-1)^{\varepsilon(\sigma)}c_{1\,\sigma(1)}c_{2\,\sigma(2)}\dotsm c_{n\,\sigma(n)}=\sum_{\sigma\in S_n}(-1)^{\varepsilon(\sigma)}1\cdot a_{\sigma(2)}\dotsm a_{\sigma(n)}^{n-1}$$
shows this determinant is homogeneous in the indeterminates $a_1, a_2, \dots,a_n$, with degree
$$1+2+\dots+(n-1)=\frac{n(n-1)}2.$$
Now Vandermonde determinant is $0$ whenever $a_i=a_j$ for some $1\le i <j .\le n$. Hence it is divisible by $a_j-a_i$. As these polynomials are pairwise coprime, Gauß's lemma ensures it is divisible by their product
$$(a_2-a_1)\dotsm(a_n-a_1)(a_3-a_2)\dotsm (a_n-a_2)\dotsm(a_n-a_{n-1}).$$
Observe this product has precisely degree $\dfrac{n(n-1)}2$. Hence it is a constant × the determinant. Considering the term $a_1^{n-1}a_2^{n-2}\dotsm a_{n-1}$ in the expansion of the general formula, we see this coefficient is $1$.
