Schur evaluation at all ones ($s_{\lambda }(1^N)={\prod }_{i<j}\frac{{\lambda }_i-{\lambda }_j+j-i}{j-i}$)

Put $b_i={\lambda }_i+N-i$. In the bialternant formula, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ and compare the first nonzero determinant coefficients as $t\to 0$. They arise from the distinct degrees $0,\dots ,N-1$ in the exponential expansions. The common factor in the $y_j$ and the factorials cancel, leaving the product of differences of the $b_i$ divided by the differences of $N-i$. This proves the displayed integral specialization without dividing by a vanishing Vandermonde value.