Sum of product of products I was working on a proof, and found that, for my proof to hold, the following must be true:
Let $\Lambda_{i} > 0$ for $i \in \{0, 1, .. , k\}$, such that $\Lambda_{i} = \Lambda_{j} \iff i = j$. Then:
$$\sum_{m = 0}^{k-1} \prod_{i = 1, \neq m}^{k}\prod_{j = 0, \neq m}^{i-1}(\Lambda_j-\Lambda_i) = \prod_{i = 1}^{k-1}\prod_{j = 0}^{i-1}(\Lambda_j-\Lambda_i)$$.
I worked a couple of specific examples, where $k = 3$ and $k = 4$, enough to prove it in those cases. But in each case, I had to work through term by term. I tried induction, and ended up with a spaghetti mess of algebra. Does this expression hold for $k \in \mathbb{Z^+}$? If it doesn't, why doesn't it seem to hold for lower values?
 A: Hint: Setting $k=3$, the left-hand side gives
\begin{align*}
\color{blue}{\sum_{m = 0}^{2}}
&\color{blue}{ \prod_{\substack{i = 1\\i\ne m}}^{3}\prod_{\substack{j = 0\\j \ne m}}^{i-1}(\Lambda_j-\Lambda_i)}\\
&=\prod_{\substack{i = 1\\i\ne 0}}^{3}\prod_{\substack{j = 0\\j \ne 0}}^{i-1}(\Lambda_j-\Lambda_i)
+\prod_{\substack{i = 1\\i\ne 1}}^{3}\prod_{\substack{j = 0\\j \ne 1}}^{i-1}(\Lambda_j-\Lambda_i)
\prod_{\substack{i = 1\\i\ne 2}}^{3}\prod_{\substack{j = 0\\j \ne 2}}^{i-1}(\Lambda_j-\Lambda_i)\\
&=\prod_{i=1}^{3}\prod_{j=1}^{i-1}(\Lambda_j-\Lambda_i)
+\prod_{\substack{i = 1\\i\ne 1}}^{3}\prod_{\substack{j = 0\\j \ne 1}}^{i-1}(\Lambda_j-\Lambda_i)
\prod_{\substack{i = 1\\i\ne 2}}^{3}\prod_{\substack{j = 0\\j \ne 2}}^{i-1}(\Lambda_j-\Lambda_i)\\
&=\left(\Lambda_1-\Lambda_2\right)\left(\Lambda_1-\Lambda_3\right)\left(\Lambda_2-\Lambda_3\right)\\
&\qquad+\left(\Lambda_0-\Lambda_2\right)\left(\Lambda_0-\Lambda_3\right)\left(\Lambda_2-\Lambda_3\right)\\
&\qquad+\left(\Lambda_0-\Lambda_1\right)\left(\Lambda_0-\Lambda_3\right)\left(\Lambda_1-\Lambda_3\right)\\
&=\Lambda_3^2\left[\left(\Lambda_1-\Lambda_2\right)
+\left(\Lambda_0-\Lambda_
2\right)+\left(\Lambda_0-\Lambda_1\right)\right]+\cdots\\
&\,\,\color{blue}{=2\Lambda_3^2\left[\Lambda_0-\Lambda_2\right]+\cdots}
\end{align*}
It seems, the LHS contains terms with $\Lambda_3$ contrary to the RHS.
