From this Approximation method for the Basel Problem, these are equivalent
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = 1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}$$
Question
There are many Different ways to prove $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$ (the Basel problem) but are there similar methods to evaluate this series without using the known Basel value $\frac{\pi^2}{6}$ to avoid any circular reasoning?
$$1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} = \frac{\pi^2}{6}$$
If true, the Basel Series follows by using the transitive property of equality
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = 1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}$$
and $$1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} = \frac{\pi^2}{6}$$
then it follows that
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
