By the way, from my equation, the place to start is by swapping the sigmas:
$\begin{align*}\displaystyle \sum_{n\ge 1} \ln\left(1+\dfrac{1}{n^p}\right) & = \sum_{n\ge 1}\left[\sum_{k\ge 1} \dfrac{(-1)^{k+1}}{k}\dfrac{1}{n^{kp}}\right] \\ & = \sum_{k\ge 1}\left[\dfrac{(-1)^{k+1}}{k}\sum_{n\ge 1}\dfrac{1}{(n^k)^p}\right]\end{align*}$
Now, you have an alternating series. By the alternating series test, it will converge if every term converges and tends towards zero. Consider the inner summation. If $k=1$, this is a p-series. It converges when $p>1$. So, that is the initial limitation. Next, for any $k>1$, $\dfrac{1}{(n^k)^p} < \dfrac{1}{n^p}$, so by comparison test, if $p>1$, you have convergence for all $k$. Since $\displaystyle \sum_{n\ge 1}\dfrac{1}{n^{kp}}$ is decreasing, $\displaystyle \dfrac{1}{k}\sum_{n\ge 1} \dfrac{1}{n^{kp}} \to 0$ as $k \to \infty$. This proves that it is convergent for any $p>1$.