Irrational to power of itself is natural

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 $\sqrt{a}={\mathrm{log}}_{n}b$ 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 :)

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 $\sqrt{a}={\mathrm{log}}_{n}b$ 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 :)