Show that $\sum_{n=1}^\infty \frac{1}{(\log(n))^p}$ diverges

Question

I want to show for all $p \in [1, \infty)$, that $$\sum_{n=1}^\infty \frac{1}{(\log(n))^p}$$ diverges.

For $p = 1$ it is just: $$ \exp(n) \ge n \\ \Rightarrow \qquad n \ge \log(n) \\ \Rightarrow \frac{1}{n} \le \frac{1}{\log(n)}$$ and $$\sum_{n=1}^\infty \frac{1}{n}$$ diverges.

For $p > 1$ I have the feeling, that the following equation is true $$\forall p \in [1, \infty) : \exists n_0 : \forall n \ge n_0 : n \ge \log^p(n)$$ Which would immediately answer the question about divergence, but I cannot prove it.

Answer 1

$$\lim_{x \to \infty} \frac{\log x}{p \log \log x}= \lim_{x \to \infty} \frac{\frac1{x}}{p \frac{\frac1{x}}{\log x}} = \lim_{x \to \infty} \frac1{p} \log x = \infty$$

This shows that there is $N$ such that for all $n \ge N$, $\log n \ge p \log \log n$, i.e. $n \ge \log^p n$.

Answer 2

For the divergence of the above series use the Cauchy condensation test.