How could I prove that
$$\sum \_{m=v}^n \left(\left(\prod \_{k=v}^{m-1} \frac{k^2}{m^2-k^2}\right)\left(\prod \_{k=m+1}^n \frac{k^2}{k^2-m^2}\right)(-1)^{m-v}\right)=1$$
or, simplified,
$$\sum \_{m=v}^n \prod \_{k=v, k \neq m}^{n} \frac{k^2}{k^2-m^2}=1$$


for any positive integers $v$ and $n$, $v \leq n$? I feel this could be somehow related to binomial coefficient identities.


**Why I want it to be true?**


I got this problem while generating formula for eigenvalues of matrix of special type. I noticed that this
$$\sum \_{m=u}^n \frac{2(-1)^{m-1}(n!)^2}{m^2(n-m)!(n+m)!} \frac{m (m+u-1)!}{u (2u-1)! (m-u)!}(-4)^{u-1}$$
can be simplified to this
$$\frac{2((u-1)!)^24^{u-1}}{(2u)!}$$



An identity involving sum of probably binomial coefficients
Consider the contour integral of 
$$ 
\frac{1}{z} \prod\_{k=v}^{n} \frac{k^2}{k^2-z^2} 
$$ 
over a circle of large radius centered at $0$. Since the integrand is 
small as $|z|\to \infty$ the answer must go to zero as the radius goes to infinity. 
But inside the circle there are poles at $z=0$ and $z= \pm k$ for $k$ from $v$ to $n$. Computing the residues here gives your identity. 


Edit in response to OP's comment: The proof above uses complex analysis and 
the residue theorem; consult any introductory book in that subject. Alternatively, 
note that if $P(x)$ is a polynomial of degree $n$ with distinct roots $r\_1$, $\ldots$, $r\_n$ then 
$$ 
Q(x)=\sum\_{j=1}^{n} \frac{1}{P^{\prime}(r\_j)} \frac{P(x)}{(x-r\_j)} 
$$ 
is a polynomial of degree $n-1$ which also equals $1$ for all the $n$-points $x=r\_j$. Therefore $Q(x)$ is identically $1$. Your identity follows by taking $P(x) = \prod\_{k=v}^{n} (k^2-x^2)$, and evaluating $Q(x)=1$ at $x=0$. 
The general identity $Q(x)=1$ mentioned above is classical, and was discussed on MSE: 
see <https://math.stackexchange.com/questions/104262/sum-of-reciprocals-of-derivative-of-polynomial-at-its-roots>


