Let $n>1$ be natural number and $a_0,a_1,\dots,a_{n-1}$ be the sequence of parwise different real numbers. Also let $A_k=\{0,1,\dots,n-1\}\setminus \{k\}$. I'm looking for a simple proof of following identity:
$$\sum_{k=0}^{n-1}\frac{1}{\prod_{m\in A_k}(a_m-a_k)}=0$$
I've been able to show this by using induction over $n$, reducing the sum to common denominator and then considering numerator as polynomial in $a_{n-1}$ and showing that it is equal to $0$ in $n$ points and it's degree is at most $n-1$. However I believe there has to be some more elementary proof of this identity.
So my questions are: is there simpler proof of this identity, are there similar identities known in the literature?
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Jakub Pawlak
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2Are you familiar with Lagrange interpolation? If yes, Hint: View the LHS as the coefficient of a certain polynomial. Show that the degree of the polynomial is low enough that we can conclude the coefficient is 0. – Calvin Lin Nov 24 '23 at 17:10
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1See https://math.stackexchange.com/q/2804888/42969 – Martin R Nov 24 '23 at 17:20