 Solution

Set $L_i={\lambda }_i+k-i$, for $1\le i\le k$. These are distinct nonnegative integers with $L_1>\cdots >L_k$. Using the falling factorial, the matrix entry can be writtenWhen $k-j>L_i$, the falling factorial vanishes, agreeing with the stipulated convention for a negative factorial argument.

The column polynomials $(X{)}_{k-1},(X{)}_{k-2},\dots ,(X{)}_0$ are monic of successive descending degrees. Subtracting lower-degree combinations of columns turns them into $X^{k-1},X^{k-2},\dots ,1$ without changing the determinant. The Vandermonde determinant therefore givesThe 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 preciselyTo 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 $L_i-(k-i)={\lambda }_i$.

ConsequentlyCombining the two identities with the hook-length formula proves the determinantal dimension formula