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Everything is in $\mathbb{Z}$. Let $v_1 < v_2 < ... < v_n = k$, and $v_1 = 1$ for $k >> n$. Let $ P = \Pi_{i < j} (v_j - v_i)$. How can I show that $P \le k^{n^2}$?

There are $n + (n-1) + ... + 1 = \frac{n(n+1)}{2}$ terms in the product. Starting from $v_n - v_{n-1} = 1$, etc. Clearly, $P = 1(1*2)(1*2*3) ...(k-1)! = \Pi_{i=1}^{i=k-1} i!$. But I'm not sure about this superfactorial(?).

Also, I noticed that the product $P$ is very similar to the determinant of a Vandermonde matrix.

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    This problem has been moved to a higher level. See http://math.stackexchange.com/questions/15366/2010-12-24

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First note that there are a total of $1 + 2 + \cdots + (n-1)$ terms i.e. a total of $\frac{n(n-1)}{2}$ terms.

And $\forall n \in \mathbb{N}$, $\frac{n(n-1)}{2} < n^2$.

And $|v_i - v_j| < k$, $\forall i,j$ since $v_{i} \in [1,k]$.

Hence, we get $\displaystyle \prod_{i.

The last inequality follows from the fact that $\frac{n(n-1)}{2} < n^2$ and $k>1$

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    I tried the naive thing of equally distributing the v's and calculated the specific result, but the product is higher if you slide them toward the end (just by experiment with n=5).2010-12-23