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]\ \big(-\sum_{n\geq 1} \frac{J_k(n)}{n} \ln(1-x^n)\big) &= \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}\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.
