 Determinant formula for Specht module dimension

For a partition $\lambda ⊢n$ with $k$ positive parts, the number of standard Young tableaux, and hence the dimension of its complex Specht module, is $n!\det (1/({\lambda }_i-i+j)!)$. Negative factorial arguments give zero entries. Setting $L_i={\lambda }_i+k-i$ turns the determinant into ${\prod }_{i<j}(L_i-L_j)/{\prod }_i{L}_i!$ by the Vandermonde determinant, matching the hook-length formula.