I've been thinking about a natural number like $n$ so that $x^x=n$ for some irrational $x$ but i couldn't find anything. As i didn't know how to approach the problem at all, i tried to make some simpler cases first ($n$ is a natural number):

  1. $\sqrt a^{\sqrt a}=n$ for irrational $\sqrt a$
    My approach: $\sqrt a^{\sqrt a}=n\Rightarrow \sqrt a^a=n^{\sqrt a}$. And here, We suppose $a$ is even. This means the LHS would be a natural number, and thus $n^\sqrt a$ is natural too. This means $\sqrt a = \log_n{b}$ which i think can not be true when $b$ is not a power of $n$ because that time i think the logarithm would be transcendental (i'm not sure). But this only disproves the case for $a$ being even! Still if we can prove theres no such $n$ and $a$, we should check the next case.
  2. $a^a=n$ for algebraic and irrational $a$
    This seems more likely than the first case, but still i have no idea in approaching it.
  3. $a^a=n$ for irrational $a$
    This is indeed more general than the first two cases and we should check it if the first two cases failed! Although it may be great to find all kinds of irrational $a$s so that the equality will hold for natural $n$.

    I would appreciate any help :)

    Edit: Having Michael Hardy's answer and TomOldfield's proof for $x$ being natural OR irrational, i would still appreciate if we had something to say about $x$ being in any of these $3$ cases.
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    $\begingroup$ Since $x^x$ is continuous, $x^x=n$ has a solution for any positive natural $n$. $\endgroup$ – coffeemath Jul 9 '14 at 20:07
  • $\begingroup$ $x^x=n\implies x=e^{W(\log{n})}$ $\endgroup$ – Silynn Jul 9 '14 at 20:08
  • $\begingroup$ @coffeemath Thanks, but can we prove that it is irrational? $\endgroup$ – CODE Jul 9 '14 at 20:11
  • $\begingroup$ And also the first case is preferred anyway :) $\endgroup$ – CODE Jul 9 '14 at 20:12
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    $\begingroup$ @coffeemath Ofcourse, i mean other than the obvious cases, we must prove the found $x$ isnt rational so it would meet our requirements in case 3 :) $\endgroup$ – CODE Jul 9 '14 at 20:17

If the problem is finding a real number $x$ such that $x^x=n$, for a given $n\in \mathbb N$, then notice that the function $x\mapsto x^x$ is continuous and goes from $1$ down to a positive number less than $1$ and then up to $\infty$ as $x$ goes from $0$ to $\infty$, so the intermediate value theorem tells us a solution exists.

Finding the solution probably has to be done numerically rather than algebraically.

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    $\begingroup$ Well actually the problem is to to an irrational x so that the equality holds... So i would need an elementary proof for that as well :) $\endgroup$ – CODE Jul 9 '14 at 20:19
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    $\begingroup$ @CODE Well, suppose $(p/q)^{(p/q)}=n$ with $(p,q) = 1$. Then $p^p=n^qq^p$, so any prime that divides $q$ divides $p$. $(p,q) = 1 \implies q = 1$ i.e. $(p/q)$ is itself an integer. Thus any natural not "hit" by being $x^x$ for $x$ natural is hit by an irrational. $\endgroup$ – Tom Oldfield Jul 9 '14 at 20:26
  • $\begingroup$ @TomOldfield Dang, beat me to it. Nicely said Tom. Why don't you make that an answer? $\endgroup$ – N. Owad Jul 9 '14 at 20:32
  • $\begingroup$ @TomOldfield That was great! Now if someone could use your proof and also could say something about $x$ being in any of those 3 cases that would be a complete desired answer :) $\endgroup$ – CODE Jul 9 '14 at 20:34
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    $\begingroup$ @CODE, N.Owad, I didn't know anything about $x^x$, so wouldn't have been able to write an answer without this one. I just spotted a way to improve it a little, so Michael should feel free to add the comment to his answer if he wants to, otherwise such a short remark is fine as a comment. As to algebraicity, I'm not sure where to begin. I'll update if I have any thoughts (unlikely!). $\endgroup$ – Tom Oldfield Jul 9 '14 at 20:47

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