When this matrix is not invertible?
Solution 1:
As pointed out in many comments, this is the well-known Vandermonde matrix. It is nonsingular when all $a_i$s are distinct. We don't need determinant to see this. Suppose $c^TA=0$ for some vector $c=(c_0,c_1,\ldots,c_{n-1})^T$. Let $p(x)=c_0+c_1x+c_2x^2+\cdots+c_{n-1}x^{n-1}$. The condition $c^TA=0$ means precisely that $p(a_i)=0$ for each $i$. Since all $a_i$s are distinct, $p$ has $n$ distinct zeroes. Yet, the degree of $p$ is at most $n-1$. Therefore it must be the zero polynomial, meaning that $c=0$. Hence $A$ is nonsingular.