**I finally did it**

Hi,

I think I did it.

Actually, least common multiple of all sum members is

\(\displaystyle \prod_{i=1}^{n-1}\prod_{k=i+1}^{n}(y_i-y_k)\).

In numerator, we have the sum with member with ordinal number i is

\(\displaystyle (-1)^{i+1} \prod_{k=1}^{n-1}(y_i-z_k)\prod_{1\le k\le n-1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_k-y_r)\)

For shorter notation, let's have

\(\displaystyle \pi(i) = \prod_{1\le k\le n-1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_k-y_r)\)

When we develop the the product, we get

\(\displaystyle (y_i^{n-1} + \sum_{k=1}^{n-1}((-1)^{k}y_i^{k-1}\sum_{r=1}^{\binom{n-1}{k}}C(k,r)))\cdot \pi(i)\)

where C(k,r) is r-th combination of k elements on array \(\displaystyle (z_1,z_2,...,z_{n-1})\)

When we get back to the sum in the numerator, we get

\(\displaystyle

\sum_{i=1}^{n}y_i^{n-1}\cdot \pi(i) + \sum_{k=1}^{n-1}(-1)^{k}\sum_{r=1}^{\binom{n-1}{k}}C(k,r)\cdot \sum_{i=1}^{n}y_i^{k-1}\cdot \pi(i)\)

As we now got multipliers of z-array (their combinations) we have to prove that every

\(\displaystyle \sum_{i=1}^{n}y_i^{r}\cdot \pi(i)= \sum_{i=1}^{n}y_i^{r}\prod_{1\le k\le n-1,k \ne i}\prod_{k+1 \le r \le n, r \ne i}(y_k-y_r)=0\)

for r < n-1.

Approach used here uses the fact that all elements have the same polynomial degree and greatest degree of particular variable is n-2. Thus for each element (possible repeatable combination of y-array) we will have its negative pair as -1 is equally combined as all y-array elements.

Therefore result has to be zero.

When we look

\(\displaystyle \sum_{i=1}^{n}y_i^{n-1}\cdot \pi(i)\)

We see that it can't be zero as \(\displaystyle y_i^{n-1}\) can't be found in other sum elements where as maximal degree of particular element there is n-2. However we can look for "similar" one that has \(\displaystyle -y_l^{n-1}y_y{n-2}\) and the same products in continuation.

We then \(\displaystyle y_l^{n-2}y_i^{n-2}(y_i-y_l)(rest \; of \; product)\)

We now halved number of sum elements. This can be repeated recursively n-1 times and then we get one product that is

\(\displaystyle \prod_{i=1}^{n-1}\prod_{k=i+1}^{n}(y_i-y_k)\)

which gives what is wanted.