Let $k_1 < k_2 < ... < k_n$ will be integers. Prove that the quotient
$\frac{V_n(k_1,...,k_n)}{V_n(1,...,n)}$ is a integer.
where $V_n(x_1,...,x_n)$ is vandermonde determinant
My idea: Show that the numerator is divisible by the denominator
Can anyone give a hint or help solve it? Thank u.
Without loss of generality, we can assume that $k_1\geq n-1$ since $V_n(k_1,\ldots,k_n) = V_n(k_1+p,\ldots,k_n+p)$ for any $p\in\mathbb{Z}$.
Since $$V_n(1,\ldots,n) = 1!2!\cdots (n-1)!.$$ We have that $$\frac{V_n(k_1,\ldots,k_n)}{V_n(1,\ldots,n)} = \det\begin{pmatrix} 1 & \cdots & 1 \\ \frac{k_1}{1!} & \cdots & \frac{k_n}{1!} \\ \vdots & & \vdots \\ \frac{k_1^{n-1}}{(n-1)!} & \cdots & \frac{k_n^{n-1}}{(n-1)!} \end{pmatrix}.$$ Note that $\binom{k_i}{n-1} = \frac{k_i(k_i-1)\cdots (k_i-n+2)}{(n-1)!}$ can be regarded as a polynomial of $k_i$ with the highest order term $\frac{k_i^{n-1}}{(n-1)!}$. So we can add and subtract the proper multiples of the first $n-1$ rows to the last row to transform it into $$\binom{k_1}{n-1}\cdots \binom{k_n}{n-1}.$$ Similarly, we can transform the $(n-1)$-th row into $$ \binom{k_1}{n-2}\cdots\binom{k_n}{n-2}, $$ and $(n-2)$-th row ... Finally, the quotient equals to $$ \det\begin{pmatrix} \binom{k_1}{1} & \cdots & \binom{k_n}{1} \\ \binom{k_1}{2} & \cdots & \binom{k_n}{2} \\ \vdots & & \vdots \\ \binom{k_1}{n-1} & \cdots & \binom{k_n}{n-1} \\ \end{pmatrix}, $$ which is an integer since each entry in the matrix is integer.