High Level Determinant

We solve the recurrence relation for the coefficients of the matrix $A$ to write $\alpha(u,v) \; = \; \frac{(c+v-1)!}{(c+v-u)!} \hspace{2cm} 1 \le u,v \le c$ and hence we deduce that $\alpha(u,v+1) - \alpha(u,v) \; = \; \left\{ \begin{array}{lll} (u-1)\alpha(u-1,v) & \hspace{1cm} & 2 \le u \le c\,,\,1 \le v \le c-1 \\ 0 & & u = 1 \,,\, 1 \le v \le c-1 \end{array}\right.$ For each $1 \le K \le c$ , let $M(K)$ be the $K \times K$ matrix whose coefficients are $\alpha(u,v)$ for $1 \le u,v \le K$ . We have shown above that, if we apply suitable column operations to the matrix, that $\begin{aligned} \mathrm{det}\,M(K) & = \; \left| \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ \alpha(2,1) & \alpha(1,1) & \alpha(1,2) & \cdots & \alpha(1,K-1) \\ \alpha(3,1) & 2\alpha(2,1) & 2\alpha(2,2) & \cdots & 2\alpha(2,K-1) \\ \alpha(4,1) & 3\alpha(3,1) & 3\alpha(3,2) & \cdots & 3\alpha(3,K-1) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \alpha(K,1) & (K-1)\alpha(K-1,1) & (K-2)\alpha(K-1,2) & \cdots & (K-1)\alpha(K-1,K-1) \end{array}\right| \\ & = \left| \begin{array}{cccc} \alpha(1,1) & \alpha(1,2) & \cdots & \alpha(1,K-1) \\ 2\alpha(2,1) & 2\alpha(2,2) & \cdots & 2\alpha(2,K-1) \\ 3\alpha(3,1) & 3\alpha(3,2) & \cdots & 3\alpha(3,K-1) \\ \vdots & \vdots & \ddots & \vdots \\ (K-1)\alpha(K-1,1) & (K-2)\alpha(K-1,2) & \cdots & (K-1)\alpha(K-1,K-1) \end{array}\right| \; = \; (K-1)!\mathrm{det}\,M(K-1) \end{aligned}$ and hence $\mathrm{det}\,M(K) \; = \; \prod_{j=1}^{K-1} j!$ so that $d \; = \; |A| \; = \; \prod_{j=1}^{2016} j!$ Thus $2016!$ certainly divides $d$ , but the prime $2017$ does not divide $d$ , so that $2017!$ does not divide $d$ . Thus the answer is $\boxed{2016}$ .