Recently I have been thinking a lot about variations of the Basel Problem, and methods to solve them. Here I found the following solution to the Basel Problem by Alfredo Z. (I include the entire answer due to its brevity)
Define the following series for $ x > 0 $
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\quad.$$
Now substitute $ x = \sqrt{y}\ $ to arrive at
$$\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 1 - \frac{y}{3!}+\frac{y^2}{5!}-\frac{y^3}{7!}+\cdots\quad.$$
if we find the roots of $\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 0 $ we find that
$ y = n^2\pi^2\ $ for $ n \neq 0 $ and $ n $ in the integers
With all of this in mind, recall that for a polynomial
$ P(x) = a_{n}x^n + a_{n-1}x^{n-1} +\cdots+a_{1}x + a_{0} $ with roots $ r_{1}, r_{2}, \cdots , r_{n} $
$$\frac{1}{r_{1}} + \frac{1}{r_{2}} + \cdots + \frac{1}{r_{n}} = -\frac{a_{1}}{a_{0}}$$
Treating the above series for $ \frac{\sin \sqrt{y}\ }{\sqrt{y}\ } $ as polynomial we see that
$$\frac{1}{1^2\pi^2} + \frac{1}{2^2\pi^2} + \frac{1}{3^2\pi^2} + \cdots = -\frac{-\frac{1}{3!}}{1}$$
then multiplying both sides by $ \pi^2 $ gives the desired series.
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}$$
This solution fascinated me. Although it takes a few things for granted (such as that the Fundamental Theorem of Algebra applies to infinite polynomials) I nevertheless thought the proof was one of the most beautiful I've seen. In attempting to generalize this I stumbled across the following function and its associated power series:
$$\cos\left(\frac{x}{\sqrt{2}}\right)\cosh\left(\frac{x}{\sqrt{2}}\right) = \sum_{k=0}^\infty \frac{x^{4k}}{(4k)!}(-1)^k$$
By a simple substitution we find that
$$\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right)\cosh\left(\frac{x^{1/4}}{\sqrt{2}}\right) = \sum_{k=0}^\infty \frac{x^{k}}{(4k)!}(-1)^k$$
Which is a polynomial in $x$; noting that $\cosh$ is nowhere zero (on the real line) we solve for the roots as such:
$$\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right) = 0 \implies \frac{x^{1/4}}{\sqrt{2}} = n\pi + \frac{\pi}{2} \implies x = \frac{\pi^4(2n+1)^4}{4}$$
Using the argument for Viete's Formula in the solution above, we find that
$$\frac{4}{\pi^4}\sum_{k=0}^\infty \frac{1}{(2k+1)^4} = -\frac{-\frac{1}{4!}}{1}$$
Upon manipulation we find that
$$\color{red}{\sum_{k=0}^\infty \frac{1}{(2k+1)^4} = \frac{\pi^4}{96}}$$
From here we can bring this into a form more like that of the Basel Problem by noting that
$$\sum_{k=0}^\infty \frac{1}{(2k+2)^4} = \frac{1}{16}\sum_{k=0}^\infty \frac{1}{(k+1)^4} = \frac{1}{16}\sum_{k=1}^\infty \frac{1}{k^4}$$
Noting that this computes the even terms of $\sum_{k=1}^\infty k^{-4}$ we find that the odd terms must equal $\frac{15}{16}\sum_{k=1}^\infty k^{-4}$ which is our series calculated above
Applying this, we find that, as desired,
$$\color{red}{\sum_{k=1}^\infty \frac{1}{k^4} = \frac{\pi^4}{90}}$$
However, I have been unable to generalize this approach further. Intuition tells me that the functions desired would have power series similar to those used, and would be a combination of trigonometric functions. Nevertheless, I have only been able to solve this for the case above (which required me to switch the power series from $\sin(x)$ to $\cos(x)$ from Alfredo's proof, as neglecting $\cosh(x)$ was desirable).
My question is thus: Can this proof format be applied to solve series of the form $\sum_{k=1}^\infty k^{-n}$ for any $n$ other than $2$ and $4$?
Note: I apologize for the length of this post, but I felt that a full presentation of the proof might assist in generalizing it. If anyone has any suggestions concerning this please let me know!
Note 2: In case anyone is troubled by this seemingly naive application of Viete's Formula to infinite polynomials, know that this is perfectly valid, and is known as the Root Linear Coefficient Theorem.
To start, the "factoring" step is known as the Weierstrass factorization theorem, which asserts that you can express some functions as products of their factors.
From here, take note that you had
$$\frac{\sin(x)}x=\dots\left(1+\frac{x^2}{2^2\pi^2}\right)\left(1+\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\dots$$
which is basically what you noted.
From here, I multiply similar terms to get
$$f(x):=\frac{\sin(x)}x=\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\dots$$
and define this as $f(x)$.
using $(1-ue^{\frac{2\pi i}n})(1-ue^{\frac{4\pi i}n})\dots(1-ue^{2\pi i})=1-u^n$, we can then get
$$f(xe^{\frac{2\pi i}n})f(xe^{\frac{4\pi i}n})\dots f(xe^{2\pi i})=\left(1-\frac{x^{2n}}{1^{2n}\pi^{2n}}\right)\left(1-\frac{x^{2n}}{2^{2n}\pi^{2n}}\right)\left(1-\frac{x^{2n}}{3^{2n}\pi^{2n}}\right)\dots$$
And since this is not generalizable to $n\notin\mathbb N$, we can only solve for
$$\sum_{k=1}^\infty\frac1{k^{2n}}$$
By using this method. As for the odd values, no closed form yet exists.