How can I show that $\prod_{{n\geq1,\, n\neq k}} \left(1-\frac{k^{2}}{n^{2}}\right) = \frac{\left(-1\right)^{k-1}}{2}$? Assume $k$ positive integer. How can I show that
$$ \tag 1 \prod_{{n\geq1,\, n\neq k}} \left(1-\frac{k^{2}}{n^{2}}\right)
 = \frac{\left(-1\right)^{k-1}}{2}? $$
    I know that
$$ \tag 2 \underset{n\geq1}{\prod}\left(1-\frac{k^{2}}{n^{2}}\right)=\frac{\sin\left(\pi k\right)}{\pi k}$$
and so if $k$ is an integer the product is $0$ but how can I use these information?

If $N \ge k+1$, then 
\begin{align}
\prod_{\underset{n\neq k}{n = 1}}^N \left(1 - \frac{k^2}{n^2}\right)&= \prod_{n = 1}^{k-1}\left(1 - \frac{k^2}{n^2}\right)\prod_{n =k+ 1}^N \left(1 - \frac{k^2}{n^2}\right)\\
&= \prod_{n = 1}^{k-1} \frac{(n-k)(n+k)}{n^2} \prod_{n = k+1}^N \frac{(n-k)(n+k)}{n^2}\\
&= \frac{(-1)^{k-1}(k-1)!(k+1)\cdots (2k-1)}{[(k-1)!]^2}\cdot \frac{(N-k)!(2k+1)\cdots (N+k)}{[(k+1)(k+2)\cdots N]^2}\\ 
&= \frac{(-1)^{k-1}}{2}\cdot \frac{(N-k)!(k-1)!k(k+1)\cdots (2k-1)(2k)(2k+1)\cdots (N+k)}{[(k-1)!k(k+1)\cdots N]^2}\\
&= \frac{(-1)^{k-1}}{2}\cdot \frac{(N-k)!(N+k)!}{(N!)^2}\\
\end{align}
Since $\lim_{N\to \infty} (N-k)!(N+k)!/(N!)^2 = 1$, we deduce that $$\prod_{\underset{n\neq k}{n \geq 1}} \left(1 - \frac{k^2}{n^2}\right) = \frac{(-1)^{k-1}}{2}.$$
