can a number be written as another number to its own power 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?
 A: 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 $x$.
A: 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.
