Binomial determinants related to Hoggatt triangles Let $d$ be a positive integer and let $0\leq{k}\leq{n}$ be non-negative integers.
Let $a_d(n)=\binom{n+d-1}{d}$ and $\binom{n}{k}_{d}=\prod_{j=0}^{k-1} \frac{a_d(n-j)}{a_d(k-j)}.$
(The matrix with entries $\binom{n}{k}_{d}$ is sometimes called Hoggatt triangle of order $d$. For $d=1$ it coincides with Pascal's triangle and for $d=2$ with the triangle of Narayana numbers).
Consider the matrix $A_d(n,k)$ whose $i,j-$ entry is $\binom{n-i+j+d-1}{n-1}$ for $0\leq {i,j}\leq{k-1}.$
Computer experiments suggest that
$$\det{A_d(n,k)}=\binom{n}{k}_{d}.$$
Any idea how to prove this? Is it possible to see directly from the matrix that the determinant satisfies $\binom{n}{k}_{d}=\binom{n}{n-k}_{d}?$
 A: The Jacobi-Trudi identities give
$$s_{\lambda} = \det(h_{\lambda_i + j - i})_{i,j=1}^k$$
where $\lambda = (\lambda_1, \dots \lambda_k)$ is a partition with $k$ parts, $s_{\lambda}$ is a Schur function, and $h_i$ is a complete homogeneous symmetric function. If we work with $n$ variables and set them all equal to $1$ then
$$h_{\lambda_i + j - i}(\underbrace{1, 1, \dots 1}_{n \text{ times}}) = {\lambda_i + j - i + n - 1 \choose n - 1}$$
so comparing to the desired determinant we see that we want to set $\lambda_i = d$. (Note that it doesn't matter whether we index $i, j$ starting from $1$ or $0$ since only the difference between them matters anyway.) This gives that the desired determinant evaluation is
$$s_{d^k}(\underbrace{1, 1, \dots 1}_{n \text{ times}})$$
which is the number of semistandard Young tableaux with shape $d^k$ (a box with $d$ columns and $k$ rows) and entries in $\{ 1, \dots n \}$. These can be counted with the semistandard hook length formula, which gives
$$s_{d^k}(\underbrace{1, 1, \dots 1}_{n \text{ times}}) = \prod_{1 \le i \le k, 1 \le j \le d} \frac{n - i + j}{(k - i) + (d - j) + 1} = \prod_{1 \le i \le k} \frac{(n - i + 1)^{\overline{d}}}{(k - i + 1)^{\overline{d}}}$$
where $x^{\overline{d}}$ denotes the rising factorial $x(x + 1) \dots (x + (d-1))$. Writing $a_d(n) = \frac{n^{\overline{d}}}{d!}$ and $\frac{a_d(n-j)}{a_d(k-j)} = \frac{(n-j)^{\overline{d}}}{(k-j)^{\overline{d}}}$, and reindexing the sum above by $1$, shows that this is equivalent to the desired result.
This argument might be slight overkill and one or more of the methods in Krattenthaler's Advanced Determinant Calculus might be easier. It is maybe also worth saying that this argument can be rephrased using the Lindstrom-Gessel-Viennot lemma, since that lemma proves the Jacobi-Trudi identities and can be specialized to this case.
