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Can you provide a proof for the following claim:

$$-\displaystyle\sum_{n=1}^{\infty}\frac{J_k(n)}{n} \cdot \ln\left(1-x^n\right)=\frac{x \cdot A_{k-1}(x)}{(1-x)^k} \quad \text{for} \quad |x| < 1 \quad \text{and} \quad k>1$$ where $J_k(n)$ is the Jordan's totient function and $A_k(x)$ is the kth Eulerian polynomial.

The first few Eulerian polynomials are: $A_1(x)=1 , A_2(x)=x+1 , A_3(x)=x^2+4x+1$ , etc.

The SageMath cell that demonstrates this claim can be found here.

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The right hand side evaluates to $$\sum_{m=0}^{\infty} m^{k-1} x^m,$$ and so it remains to verify that the coefficient of $x^m$ in the l.h.s. is also $m^{k-1}$.

By differentiating the l.h.s., we have \begin{split} [x^m]\ \bigg(-\sum_{n\geq 1} \frac{J_k(n)}{n} \ln(1-x^n)\bigg) &= \frac1m [x^{m-1}]\ \sum_{n\geq 1} J_k(n)\frac{x^{n-1}}{1-x^n}\\ &=\frac1m \sum_{n|m} J_k(n) \\ &= m^{k-1} \end{split} as required.

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