Prove that $ζ(4)=π^4/90$ knowing that $\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$

Question

The question

Knowing that: $$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right) \tag{1}$$ obtain the Taylor series expansion of $\frac{\sin(πz)}{πz}$ to deduce: $$ \sum_{1 ≤ n_1 < n_2 < … < n_k} \frac{1}{n_1^2n_2^2 … n_k^2} = \frac{π^{2k}}{(2k + 1)!} \tag{2}$$ Also deduce that: $$ ζ(4) = \sum_{n=1}^∞ \frac{1}{n^4} = \frac{π^4}{90} \tag{3}$$

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What I've obtained

For (2) I've done the following (doing what the question suggests): $$ 1 - \frac{z^2π^2}{3!} + \frac{z^4π^4}{5!} - … = (1 - z^2)\left(1 - \frac{z^2}{2^2}\right)\left(1 - \frac{z^2}{3^2}\right)\left(1 - \frac{z^2}{4^2}\right)…= $$ $$ = 1 - z^2\left( 1 + \frac{1}{2^2} + \frac{1}{3^2} + … \right) + z^4 \left( \frac{1}{1^22^2} + \frac{1}{1^23^2} + … + \frac{1}{2^23^2}+…\right) - z^6\left( \frac{1}{1^22^23^2} +\frac{1}{1^22^24^2} + … + \frac{1}{1^23^24^2} + … \right) + … =$$ $$ = 1 - z^2\sum_{1 ≤ n_1}\frac{1}{n_1^2} + z^4 \sum_{1 ≤ n_1 < n_2}\frac{1}{n_1^2n_2^2} + … $$ So one sees that (2) necessarily holds.

For (3) I've tried taking $\sum_{1 ≤ n_1<n_2}\frac{1}{n_1^2n_2^2}$ in order to have exponent 4 in the numerator, but according to (2) that gives $π^4/120$. That try is obviously not correct since $n_1≠n_2$ and $90≠120$, but I don't know how how to prove it.

Any help?

Answer 1

Hint: $\frac{\pi^4}{90}=\left(\frac{\pi^2}{6}\right)^2-2\left(\frac{\pi^4}{120}\right)$.