I'm going through George Bergman's exercises for Rudin's PMA Chapter 3 (Sequences and Series) and Bergman asks the following:
Prove or disprove: For every positive real number $p$, one has $\lim_{n \to \infty}n^{n^{-p}} = 1$. (Rudin proves the case for $p = 1$).
One can see this is true using strong theorems, but...
Since this question is in Chapter 3, things not yet introduced shouldn't be used like $\log$, $e$, derivatives, etc. (although Rudin gives a rudi(n)mentary definition of $\log$ in Ch. 1 Exc. 7). The case for $p > 1$ is easy, since $n^{n^{-p}} \leq n^{n^{-1}}$.But my biggest struggle is when 0 < p < 1.
I've tried copying Rudin by trying to come up with a clever manipulation of binomial theorem, but I have not gotten any insight, since $n^p$ is not necessarily an integer, or even rational. And even when I try to brute-force my way with binomial theorem, it seems that I have to prove an alternative formulation of the same question.
I've tried thinking of an $\epsilon$-$N$ proof, since $n^{-p} \to 0$ as was proven by Rudin. But I think that will require $\log$ since $n^{n^{-p}} = b^{n^{-p}\log_b{n}},\ b > 1$. Maybe Bergman allows this if one has completed Ch. 1 Exc. 7, but he did not say that it was prerequisite.
I've tried thinking of a clever equality/inequality to use here (QM-AM-GM-HM, telescoping, binomial theorem) but the possibility that $n^p$ can be irrational and the sequence can converge extremely slowly seems to nullify the usefulness of many equations.
Perhaps I can show it for $p$, where $p$ is in the range of a sequence converging to $0$ like $\left(\frac{1}{n}\right)$. Then choose $n$ large enough and use binomial theorem?