I want to show that

\begin{align*} \sum_{\sigma \in S_{n+1}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n+1} a_{i, \sigma(i)} = \frac{1}{{a_{1, 1}^{n-1}}}\sum_{\sigma\in S_{n}}^{}{\text{sgn}\left(\sigma\right)} \prod_{i = 1}^{n} \left( a_{i+1, \sigma(i)+1}\cdot a_{1, 1}- a_{1, \sigma(i)+1}\cdot a_{i+1, 1} \right) .\end{align*} However, I'm not allowed to give an argument that involves the determinant and its properties. Could someone give me a hint how to start? (I have seen a question involving the above identity here on this platform).

  • $\begingroup$ @ancientmathematician yes, it's $\sigma(i) + 1$. What you proposed does not make much sense because $n+1 \notin S_n$ $\endgroup$
    – Richard
    Commented Feb 18, 2022 at 13:31


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