Polylog of root of unity

Lemma:

$\sum _{k=1}^n\sum _{m=1}^{\infty} \omega^{km}f_m=n\sum _{m=1}^{\infty}f_{nm}$

Proof:

Rearranging the terms give

$\sum _{k=1}^n\sum _{m=1}^{\infty} \omega^{km}f_m=\sum _{m=1}^{\infty}\sum _{k=1}^nf_m\omega^{km}=\sum _{m=1}^{\infty}f_m\sum _{k=1}^n\omega^{km}$

$\sum _{k=1}^n\omega ^{km}=\omega \frac{1-\omega ^k}{1-\omega }=\begin{cases}n \text{ if } k \text{ is a multiple of }n\\0 \text{ otherwise}\end{cases}$

Hence

$\sum _{m=1}^{\infty}f_m\sum _{k=1}^n\omega^{km}=\sum_{m=1}^{\infty}nf_{nm}=n\sum_{m=1}^{\infty}f_{nm}$

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Now for the question, we want to find

$\sum _{k=1}^n\sum _{m=1}^{\infty}\frac{\omega ^{km}x^m}{m^2}$

So, making $f_m=\frac{x^m}{m^2}$ gives

$\sum _{k=1}^n\sum _{m=1}^{\infty}\frac{\omega ^{km}x^m}{m^2}=n\sum_{m=1}^{\infty}\frac{x^{mn}}{(mn)^2}=n^{-1}\text{Li}_2(x^n)$

The answer follows

Extra

Let's compare the derivatives of both sides, with the derivative of the LHS being $-\sum\frac{\ln(1-\omega^kx)}{x}$ .

Now $\frac{d}{dx}\left(Li_2(x^n)\right)=-\frac{n\ln(1-x^n)}{x}=-\frac{n\ln(\prod(1-\omega^kx))}{x}=-n\sum\frac{\ln(1-\omega^kx)}{x}$

It follows that $f(n)=0$ and $g(n)=\frac{1}{n}$ , so that the answer is $\boxed{1}$