Proof that $\frac{V_n(k_1,...,k_n)}{V_n(1,...,n)}$ is a integer. 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.
 A: 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.
