= Standard skew tableau determinant
{title2=$f^{\lambda/\mu}=n!\det\!\left[\dfrac1{(\lambda_i-\mu_j-i+j)!}\right]$}
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>.
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