Computig the series $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$

So I have this problem for midterm reviews:

$$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$

I know that you can find the series form of a natural log, as shown here:

$$\ln\left(1-\frac{1}{n^2}\right)=-\sum_{k=2}^\infty \left(\frac{1}{n^{4k}}\right)\left(\frac{1}{2k}\right) $$

But the above doesn't seem to help very much since it results in two summation notations mushed together. Is there a somewhat nontedious way to go about this? Thanks! All help appreciated.

$$\begin{align*}\log(1-1/n^2)&=\log\left(\frac{n^2-1}{n^2}\right)\\&=\log(n^2-1)-\log(n^2)\\&=\log[(n+1)(n-1)]-2\log(n)\\&=\log(n+1)+\log(n-1)-2\log(n)\end{align*}$$... and our series telescopes. The only term that survives is $-\log 2$.$$\begin{align*}\sum_{n=2}^\infty\log(1-1/n^2)&=\sum_{n=2}^\infty\left(\log(n+1)+\log(n-1)-2\log (n)\right)\\&=\log3+\log1\color{red}{-2\log2}+\log4+\color{red}{\log2}-2\log3+\dots\\&=-\log 2\end{align*}$$

$$ \sum_{n=2}^{\infty} \ln \left( 1 - \frac{1}{n^2} \right) = \ln \left( \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2} \right) \right) $$

$$ = \ln \left( \prod_{n=2}^{\infty} \frac{n-1}{n}\frac{n+1}{n} \right) $$

$$ \ln \left( \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times ... \right) $$

$$ = \ln \left( \frac{1}{2} \right) = -\ln 2 $$