As a simpler example, consider the iterated sum with only 2 indices,
[tex]\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac1{n_1n_2(n_1+n_2)}[/tex]
(The case with just one index is pretty simple, as it reduces to ζ(2) = π²/6.)
Let
[tex]\displaystyle f(x) = \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac{x^{n_1+n_2}}{n_1n_2(n_1+n_2)}[/tex]
Differentiating and multiplying by x, we get
[tex]\displaystyle x f'(x) = \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac{x^{n_1+n_2}}{n_1n_2} \\\\ = \left(\sum_{n_1=1}^\infty\frac{x^{n_1}}{n_1}\right) \left(\sum_{n_2=1}^\infty \frac{x^{n_2}}{n_2}\right) \\\\ = (-\ln(1-x))^2 = \ln^2(1-x)[/tex]
[tex]\implies f'(x) = \dfrac{\ln^2(1-x)}x[/tex]
By the fundamental theorem of calculus (observing that letting x = 0 in the sum makes it vanish), we have
[tex]f(x) = \displaystyle \int_0^x \frac{\ln^2(1-t)}t \, dt[/tex]
If we let x approach 1 from below, f(x) will converge to the double sum and
[tex]\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \frac1{n_1n_2(n_1+n_2)} = \int_0^1 \frac{\ln^2(1-x)}x \, dx[/tex]
In the integral, substitute [tex]x\mapsto1-x[/tex], use the power series expansion for 1/(1 - x), and integrate by parts twice.
[tex]\displaystyle \int_0^1 \frac{\ln^2(1-x)}x \, dx = \int_0^1 \frac{\ln^2(x)}{1-x} \, dx \\\\ = \sum_{m=0}^\infty \int_0^1 x^m \ln^2(x) \, dx \\\\ = \sum_{m=0}^\infty -\frac2{m+1} \int_0^1 x^m \ln(x) \, dx \\\\ = \sum_{m=0}^\infty \frac2{(m+1)^2} \int_0^1 x^m \, dx \\\\ = 2 \sum_{m=0}^\infty \frac1{(m+1)^3} \\\\ = 2 \sum_{m=1}^\infty \frac1{m^3} = 2\zeta(3)[/tex]
We can generalize this method to k indices to show that
[tex]\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty \frac1{n_1n_2\cdots n_k(n_1+n_2+\cdots+n_k)} = (-1)^k \int_0^1 \frac{\ln^k(1-x)}x \, dx \\\\ = k!\,\zeta(k+1) = \Gamma(k+1)\zeta(k+1)[/tex]
Then the sum we want is
[tex]\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_{2022}=1}^\infty \frac1{n_1n_2\cdots n_{2022}(n_1+n_2+\cdots+n_{2022})} = \boxed{\Gamma(2023)\zeta(2023)}[/tex]
