Generalizing Variant Proof of Basel Problem 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.
 A: 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.
