Let $\lambda = (\lambda_1, ..., \lambda_{n})$ be a partition, $\lambda^{'}$ be the conjugate of $\lambda$, and $$A_{\lambda} := \det[\tilde{h}_{\lambda_i - i + j}^{(i)}]_{1 \leq i, j \leq n},$$ where $\tilde{h}_{p}^{(i)} = \sum_{k \geq 0}\binom{i-1}{k}(-1)^kh_{p+k}(x_1, ..., x_n)$, and $h_{p}$ is the complete symmetric polynomial with index $p$.
I was able to show that $$A_{\lambda} = \det\biggr[\sum_{k\geq 0}\binom{i-1}{k}(-1)^ke_{\lambda_i^{'} - i + j + k}\biggl(\frac{x_1}{1 - x_1}, ..., \frac{x_n}{1 - x_n}\biggr)\biggl]_{1 \leq i, j \leq n}, \hspace{10mm} (1)$$ where $e_{p}$ is the elementary symmetric polynomial with index $p$.
Somewhere I found that the following is also true $$A_{\lambda} = \det\biggr[\sum_{k \geq 0}\binom{\lambda_i^{'}-1+k}{k}(-1)^ke_{\lambda_i^{'} - i + j + k}(x_1, ..., x_n)\biggl]_{1 \leq i, j \leq \ell(\lambda^{'})}, \hspace{10mm} (2)$$ where $\ell(\lambda^{'})$ is the length of $\lambda^{'}$. I have been trying to get $(2)$ from $(1)$ but I have failed. I think I'm forgetting an important determinant or binomial coefficient property in order to get $(2)$.
Any help will be really appreciated!