Finding determinant of matrix involving binomial coefficients
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 $$ \frac{1}{n+1} \left( \begin{array}{c} q+n-1 \\ n \end{array} \right) \left( \begin{array}{c} q+n-2 \\ n \end{array} \right) $$
? det7 = matdet(d)
%26 = 1/3628800*q^12 + 1/151200*q^11 + 7/103680*q^10 + 23/60480*q^9 + 1541/1209600*q^8 + 1/400*q^7 + 1747/725760*q^6 - 13/60480*q^5 - 383/129600*q^4 - 101/37800*q^3 - 1/1260*q^2
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? factor(det4 * 144)
%16 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 1]
? factor(det5 * 2880)
%15 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 1]
? factor(det6 * 86400)
%20 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 1]
?
? factor(det7 * 3628800)
%27 =
[q - 1 1]
[ q 2]
[q + 1 2]
[q + 2 2]
[q + 3 2]
[q + 4 2]
[q + 5 1]
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? factor(144)
%21 =
[2 4]
[3 2]
? factor(2880)
%22 =
[2 6]
[3 2]
[5 1]
? factor(86400)
%23 =
[2 7]
[3 3]
[5 2]
?
? factor( 3628800)
%28 =
[2 8]
[3 4]
[5 2]
[7 1]