Suppose you have a sequence of distinct real coefficients $\lambda_1, \lambda_2, \lambda_3,...$ and define the function:
$$H(r) \equiv \sum_{k=1}^r \coprod_{\ell = 1, \ell \neq k}^r (\lambda_k - \lambda_\ell) \quad \quad \quad \quad \quad \text{for } r \in \mathbb{N}.$$
(By convention, the empty product is taken to be unity, so we have $H(1)=1$ as the first value of this function.) I would like to establish that $H(r) = 0$ for all $r>1$, but I am having trouble proving this. When I expand out for the first few values I can establish the result for the individual cases, and this involves turning each inverted polynomial into a product over all combinations of distinct coefficients. However, I have not been able to prove the general case.
Questions: How do you prove this result? Is this result a well-known polynomial result (or related to a well-known result)?
My working so far: It is fairly simple to establish the result for individual values by expanding and collecting like terms. For $r=2$ we have:
$$\begin{align} H(2) &= \coprod_{\ell = 1, \ell \neq 1}^2 (\lambda_1 - \lambda_\ell) + \coprod_{\ell = 1, \ell \neq 2}^2 (\lambda_2 - \lambda_\ell) \\[6pt] &= \frac{1}{\lambda_1 - \lambda_2} + \frac{1}{\lambda_2 - \lambda_1} \\[6pt] &= \frac{1}{\lambda_1 - \lambda_2} - \frac{1}{\lambda_1 - \lambda_2} = 0. \\[6pt] \end{align}$$
For $r=3$ we have:
$$\begin{align} H(3) &= \coprod_{\ell = 1, \ell \neq 1}^3 (\lambda_1 - \lambda_\ell) + \coprod_{\ell = 1, \ell \neq 2}^3 (\lambda_2 - \lambda_\ell) + \coprod_{\ell = 1, \ell \neq 3}^3 (\lambda_3 - \lambda_\ell) \\[6pt] &= \frac{1}{(\lambda_1 - \lambda_2)(\lambda_1 - \lambda_3)} + \frac{1}{(\lambda_2 - \lambda_1)(\lambda_2 - \lambda_3)} + \frac{1}{(\lambda_3 - \lambda_1)(\lambda_3 - \lambda_2)} \\[6pt] &= \frac{(\lambda_2 - \lambda_3)}{(\lambda_1 - \lambda_2)(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3)} - \frac{(\lambda_1 - \lambda_3)}{(\lambda_1 - \lambda_2)(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3)} + \frac{(\lambda_1 - \lambda_2)}{(\lambda_1 - \lambda_2)(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3)} \\[6pt] &= \frac{(\lambda_2 - \lambda_3)-(\lambda_1 - \lambda_3)+(\lambda_1 - \lambda_2)}{(\lambda_1 - \lambda_2)(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3)} = 0. \\[6pt] \end{align}$$
It is possible to go higher, but the algebra becomes cumbersome. The method seems to be to write each term with a denominator that is a product of all distinct pairs, and then the numerators cancel out. However, I have not been able to prove this for the general case.
Just double-checking - when putting into a single fraction, the numerator is
\begin{align} \sum_{i=1}^r (-1)^{i-1} \Pi_{1 \leq k < l \leq r, k \neq i, l \neq i} (\lambda_k-\lambda_l) \text{ ?} \end{align}