can a number be written as another number to its own power

Question

Do we know which numbers can be written as another number to its own power? More precisely, if we have a number $x$, when can we write it as $n^n$ for some $n\in \mathbb{N}$?

For example, if $q$ is a prime at least $5$ and $x= (q(q^2-1))^{q^2}$, can we write $x=n^n$ for some $n\in \mathbb{N}$?

Intuition says that writing $n^n$ in its prime factorization, we have each of those primes divides the power by definition. Whereas with $x$, $q$ divides $q^2$ but $q^2-1$ does not divide it, does any of this make sense?

Answer 1

We want to argue that the given $x$ can not be of the form $n^n$.

For any prime $p$ and a natural number $m$, let $v_p(m)$ denote the order to which $p$ divides $m$. Thus $v_2(24)=3$ since $2^3\,|\,24$ but $2^4\,\nmid \,24$. for example.

Suppose that $q$ is a prime and that $$x=\left(q(q^2-1)\right)^{q^2}=n^n$$ for some natural number $n$. We will derive a contradiction.

Since $q\,\nmid\,(q^2-1)$ we easily see that $v_q(x)=q^2$. It is also easy to see that $v_q(n^n)=v_q(n)\times n$.

Let $v_q(n)=a$. Then $n=q^am$ with $q\,\nmid\,m$. And we see that $$q^2=v_q(x)=v_q(n^n)=a\times q^a\times m$$

But this is impossible. It would imply that $m=1$ since otherwise we'd have $m\,|\,q^2$ which is impossible for $m>1$ (as $m$ is prime to $q$ by construction.). But if $m=1$ then $n=q^a$ which is impossible since $\gcd(q^2-1,n)>1$. And we are done.

Answer 2

By your definition, the list of numbers for which that can be done are $$1^1, 2^2, 3^3, 4^4, 5^5, 6^6...$$

If you broaden it to include non integers, the answer becomes a little more interesting, as the graph of $x^x$ has a minimum at $(1/e, (1/e)^{1/e}) \approx (0.368,0.692)$, so any number $y$ higher than about $0.692$ can be written as $y = x^x$ for some $x$.