TITLE: Show that $\Pi_{i < j} (v_i - v_j) \le k^{n^2}$ for $1 \le v_1 < v_2 < ... < v_n = k$
QUESTION [0 upvotes]: 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.

REPLY [2 votes]: 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 $\displaystyle \prod_{i<j} |v_i - v_j| < k^{\frac{n(n-1)}{2}} < k^{n^2}$.
The last inequality follows from the fact that $\frac{n(n-1)}{2} < n^2$ and $k>1$
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