= Solution
Set $L_i=\lambda_i+k-i$, for $1\leq i\leq k$. These are distinct nonnegative integers with $L_1>\cdots>L_k$. Using the <falling factorial>, the matrix entry can be written
$$
\frac1{(\lambda_i-i+j)!}
=\frac{(L_i)_{k-j}}{L_i!}.
$$
When $k-j>L_i$, the falling factorial vanishes, agreeing with the stipulated convention for a negative factorial argument.
The column <polynomials> $(X)_{k-1},(X)_{k-2},\ldots,(X)_0$ are monic of successive descending degrees. Subtracting lower-degree combinations of columns turns them into $X^{k-1},X^{k-2},\ldots,1$ without changing the <determinant>. The <Vandermonde determinant> therefore gives
$$
\det\left(\frac1{(\lambda_i-i+j)!}\right)
=\frac{\prod_{i<j}(L_i-L_j)}{\prod_iL_i!}.
$$
The positive sign comes from reversing the usual increasing-degree Vandermonde columns at the same time as writing the differences as $L_i-L_j$.
We now match this denominator with the <hook-length formula>. The <hook lengths> in row $i$ are precisely
$$
\{1,2,\ldots,L_i\}\setminus\{L_i-L_r:r>i\}.
$$
To verify the claim, the first hook in the row has length $L_i$. As the column index passes from $j$ to $j+1$, its <hook length> drops by $1$ plus the number of lower rows ending at column $j$. The skipped lengths are exactly $L_i-L_r$ for those rows: if $\lambda_r=j$, then $L_i-L_r=\lambda_i-j+r-i$. At the right-hand end, lower rows having the same length as row $i$ account for the missing smallest lengths by the same formula. This describes all omitted values, and the remaining count is $L_i-(k-i)=\lambda_i$.
Consequently
$$
\prod_{x\in\lambda}h(x)
=\prod_i\frac{L_i!}{\prod_{r>i}(L_i-L_r)}
=\frac{\prod_iL_i!}{\prod_{i<j}(L_i-L_j)}.
$$
Combining the two identities with the <hook-length formula> proves \b[the determinantal dimension formula]
$$
\boxed{f_\lambda=n!\det\left(\frac1{(\lambda_i-i+j)!}\right)}.
$$
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