Standard skew tableau determinant ($f^{\lambda /\mu }=n!\det \left[\frac{1}{({\lambda }_i-{\mu }_j-i+j)!}\right]$)

Here $n=|\lambda |-|\mu |$, with zero entries for negative factorial arguments. Extract the squarefree monomial $x_1\cdots x_n$ from the skew Jacobi–Trudi identity. A product $h_{r_1}\cdots h_{r_{\ell }}$ contributes $n!/\prod r_i!$ because its distinct labels are assigned to the factors in that many ways. Summing determinant signs proves the formula, since a squarefree semistandard Young tableau is a standard skew Young tableau.