I have been trying to solve a problem and have given my answer in the form of a determinant, namely: \begin{align*} \det\begin{pmatrix} \binom{q}{2} & q & 1 & 0 & \dots & 0 \\ \binom{q}{3} & \binom{q}{2} & q & 1 & \dots & 0 \\ \binom{q}{4} & \binom{q}{3} & \binom{q}{2} & q & \dots & 0 \\ \binom{q}{5} & \binom{q}{4} & \binom{q}{3} & \binom{q}{2} & \dots & 0 \\ \vdots & \vdots & \vdots & \dots & \ddots & q \\ \binom{q}{n+1} & \binom{q}{n} & \binom{q}{n-1} & \binom{q}{n-2} & \dots & \binom{q}{2} \end{pmatrix} \end{align*} This is probably enough, but I thought I'd challange myself and dust off my old linear algebra skills. Unfortunately, it is not going well. I have attempted brute forcing it by multiplying out coefficients, but I don't think that is the way to go, because I am getting nowhere. Everytime I touch it I seem to be increasing the complexity rather than decreasing it...
Is there a good way to tackle this problem, or is this as good as it gets?
Any help would be much appreciated!
seems quite rigid pattern; here my number would be your $n+1.$ I put in the one with 7 so you could see the coefficient of the highest degree term. I think the constants factor in a similar way, as $$ 144 = 1 \cdot 2^2 \cdot 3^2 \cdot 4, $$ $$ 2880 = 1 \cdot 2^2 \cdot 3^2 \cdot 4^2 \cdot 5, $$ $$ 86400 = 1 \cdot 2^2 \cdot 3^2 \cdot 4^2 \cdot 5^2 \cdot 6, $$ $$ 3628800 = 1 \cdot 2^2 \cdot 3^2 \cdot 4^2 \cdot 5^2 \cdot 6^2 \cdot 7, $$
That leads to a solid recursion, let me call my number $w,$ $$ d_{w+1} = \left( \frac{(q+w-2)(q+w-3)}{w(w-1)} \right) \; d_w $$ See if you can write that securely in your notation.
One thing that comes out is a closed form. If the lower left element has $n+1,$ the determinant comes out $$ \frac{1}{n+1} \left( \begin{array}{c} q+n-1 \\ n \end{array} \right) \left( \begin{array}{c} q+n-2 \\ n \end{array} \right) $$