Suppose we have the polynomial $$P(z)=\sum_{k=0}^n a_k z^k$$ where $a_k,z\in\mathbb{C}$ such that $a_n\ne 0$. If this polynomial has roots $r_1,\ldots,r_n$, then by Vieta's formulas $$\sum_{k=1}^n \frac{1}{r_k}=-\frac{a_1}{a_0}.$$
Now $$\frac{\sin z}{z}=\sum_{k=0}^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}$$ has the roots $z=\pm \pi, \pm 2\pi,\ldots$, so, upon setting $y=z^2$, the equation $$\sum_{k=0}^\infty (-1)^k \frac{y^k}{(2k+1)!}=0$$ has the solutions $y=\pi^2, 4\pi^2,9\pi^2,\ldots$.
If we treat the left-hand side of this equation as a polynomial and apply Vieta's formula we get $$\sum_{k=1}^\infty \frac{1}{k^2\pi^2}=\frac{1}{3!}\implies \sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$
Question: Can we rigorously prove, without using $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$, that the "sine series" "obeys" Vieta's formula? Does such a theorem hold more generally for other power series of complex functions? If it does, we would have an elegant proof of $\sum_{k=1}^\infty \frac{1}{k^2}= \frac{\pi^2}{6}$.
Edit: As mentioned by Ethan Bolker, I'm trying to rigorously justify Euler's first method of solving the Basel problem.
Assume $g$ is an entire function of the form $g(z)=1+\sum_{n \ge 1} a_nz^n=\Pi_{k \ge 1} (1-z/z_k)$ where $\sum |1/z_k| < \infty$.
Then we claim that $a_1=-\sum_{k \ge 1}\frac{1}{z_k}$ and more generally $(-1)^na_n=c_n((z_k))=\lim_{m \to \infty}c_n(z_k,m)$ where $c_m(z_k,n)$ is the $n$ Viete sum of the polynomial $P_m(z)=\Pi_{k \le m} (1-z/z_k)$ (in other words for $m \ge n$ we take $c_m(z_k,n)$ the sums of $n$ products of distinct $1/z_k, k=1,..m$ which are $(-1)^n$ times the $z^n$ coefficient of $P_m$
The proof is immediate since by hypothesis $P_m \to f$ normally on compact sets, so in particular $$\int_{|z|=1}P_m(z)z^{-n-1}dz \to \int_{|z|=1}f(z)z^{-n-1}dz, m \to \infty$$
But by direct computation (or residues) we have that the term on the left-hand side above is $2\pi i$ times the coefficient of $z^n$ in $P_m$, hence by definition, it is $2\pi i (-1)^nc_n(z_k,m)$, while the term on the right hand side is $2\pi i a_n$
For $n=1$ we recover our claim since $c_1(z_k,m)=\sum_{k \le m}1/z_k$
Applying the above to $\Pi_{k \ge m} (1-z/k^2)=\frac{\sin \pi \sqrt z}{\pi \sqrt z}=\sum_{k \ge 0}\frac{(-1)^k\pi^{2k}z^k}{(2k+1)!}=1-\pi^2z^2/6+...$ we recover the required result.
Note that inductively we can compute $\zeta(2n)$ this way since, with the notation above, $\zeta(4)=\zeta(2)^2-2c_2((1/k^2))$ and we know that $c_2=a_2=\pi^4/120$ hence $\zeta(4)=\pi^4/36-\pi^4/60=\pi^4/90$ and so on