Ideal of distinct points Suppose we work on $k[x_1,\ldots,x_n]$ ($k$ is algebraically closed and char $k=0$). I want to prove that the ideal 
$$I=(x^{a_n+1}_n-x^{a_n+1}_1,\ldots,x^{a_2+1}_2-x^{a_2+1}_1)$$
with $1\leq a_1\leq\ldots\leq a_n \in \mathbb{N}$ defines $\prod_{i=2}^n (a_i+1)$ distinct points in $\mathbb{P}^{n-1}$. I did the simplest cases ($n=2,3$, $a_i=2,3$), but I can't find a general formula per the primary decomposition of $I$ (although I'm pretty sure there exist one). 
Can someone help me, finding this formula or maybe giving me a reference? Thanks in advance.
 A: This is not quite true as stated: the number of points should be $\prod_{i=2}^n (a_i+1)$ and one should require that the characteristic of $k$ not divide any $a_i+1$.
If $x_1 = 0$, then $x_i=0$ for all $i\leq n$, as each equation becomes $x_i^{a_i+1}=0$. So we may assume that $x_1\neq 0$, and thus $V(I)$ is entirely contained in $D(x_1)$. So we may dehomogenize with respect to $x_1$ by setting $t_i=\frac{x_i}{x_1}$ and work on the affine problem of determining what $V(t_n^{a_n+1}-1,t_{n-1}^{a_{n-1}+1}-1,\cdots,t_2^{a_2+1}-1)$ is. As $k$ is algebraically closed and $\operatorname{char} k$ does not divide $a_i+1$ for any $i$, each equation has $a_i+1$ distinct solutions (here we use that the polynomial $t_i^{a_i+1}-1$ is separable). So there are $\prod_{i=2}^n (a_i+1)$ distinct solutions to the entire system.
When $\operatorname{char}k=p$ does divide $a_i+1$, one removes $a_i+1$ from the product and replaces it with $\frac{a_i+1}{p^{b_i}}$ where $b_i$ is the maximal power of $p$ which divides $a_i+1$. (If you're working from the scheme viewpoint, these turn in to fat points, so in some sense the right number of points are still there.)
