Why does summation of infinite series end up in powers of pi?

Question

For instance, we have $$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...=\frac{\pi}{4}$$ $$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}$$ $$1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+...=\frac{\pi^3}{32}$$ $$1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+...=\frac{\pi^4}{90}$$

All the $\pi$s come up for no apparent reason. Is there any reason why infinite series give rise to $\pi$, especially in the case when it ends up in powers of $\pi$?

I can think of Fourier transform and Reimann Zeta function as an approach, but I'm not a math guy so I have no idea how to explain this.

Edit: Further powers: $$1-\frac{1}{3^5}+\frac{1}{5^5}-\frac{1}{7^5}+...=\frac{5\pi^5}{1536}$$ $$1+\frac{1}{2^6}+\frac{1}{3^6}+\frac{1}{4^6}+...=\frac{\pi^6}{945}$$ $$1-\frac{1}{3^7}+\frac{1}{5^7}-\frac{1}{7^7}+...=\frac{61\pi^7}{184320}$$ $$1+\frac{1}{2^8}+\frac{1}{3^8}+\frac{1}{4^8}+...=\frac{\pi^8}{9450}$$

Answer 1

One way to arrive at some such series expressions involving $\pi$ is to start from $$\tag1\frac1\pi\sin \pi z = z\prod_n(1-\frac{z^2}{n^2})$$ Of course, first of all you have to justify $(1)$; you may at least notice that the zeroes are in the right places and that naive (but justifyable) differentiation produces the same derivative at $z=0$. Next, (again: naively, but this can be justified) develop into powers of $z$, to find $$z-\frac{\pi^2}6z^3+\frac{\pi^4}{120}z^5\pm\ldots =z-z^3\sum\frac1{n^2}+z^5\prod_{n<m}\frac1{(nm)^2}\pm\ldots$$ and compare coefficients. This gives you $\sum_n \frac1{n^2}=\frac{\pi^2}6$ directly and then from $$ \sum_n\frac1{n^4} = \left(\sum_n\frac1{n^2}\right)^2 - 2\sum_{n<m}\frac1{(nm)^2}=\frac{\pi^4}{6^2}-2\cdot\frac{\pi^4}{120}=\frac{\pi^4}{90}$$ and so on