Why does $(x^n)^{x^n} = a^a$ imply $x^n = a$? I came across a problem using a property of exponents that I was not aware of. In one of the steps, an equation was simplified to this equality:
$$(x^5)^{x^5}=10^{10}$$
Knowing the solution, the above equation was simplified to:
$$x^5=10$$
What property is being used here to determine this simplification? I understand that given an equation of the form $$x^n = x^m$$ implies $$n = m$$
So, how can one determine that $$(x^n)^{x^n} = a^a$$ implies $$x^n = a$$?
Thank you.
 A: Let $x^5=y$.
Thus, since $f(y)=y^y$ increases on $\left[\frac{1}{e},+\infty\right)$ and for $0<y<1$ our equation has no roots, we obtain $x^5=10$ and $x=\sqrt[5]{10}.$
A: The general claim in the title is not true.
In the first place, $x^n$ is any non-negative real, and you can as well reason on
$$b^b=a^a.$$
A counterexample is
$$\frac12^{1/2}=\frac14^{1/4}.$$
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Consider more general form
\begin{align}
z^z&=a^a
\tag{1}\label{1}
,
\end{align}
which is equivalent to
\begin{align}
z\ln(z)&=a\ln(a)
\tag{2}\label{2}
\\
\text{or }\quad
\ln(z)\exp(\ln(z))
&=
\ln(a)\exp(\ln(a))
\tag{3}\label{3}
\quad\text{for } a>0
.
\end{align}
Clearly, \eqref{3} is
in a form $u\exp u=v\exp v$,
just a textbook equation
to solve in terms of the Lambert $\W$ function:
\begin{align}
\W\left(\ln (z)\exp(\ln (z))\right)
&=
\W\left(\ln (a)\exp(\ln(a))\right)
\tag{4}\label{4}
\\
\ln(z)&=\W\left(\ln(a)\exp(\ln(a))\right)
\tag{5}\label{5}
,\\
z&=\exp\left(\W\left(\ln(a)\exp(\ln(a))\right)\right)
\tag{6}\label{6}
.
\end{align}
The number of real solutions of \eqref{5} (and hence \eqref{1})
depends on the argument $u=\ln(a)\exp(\ln(a))$ of $\W$ on the right-hand side.
If $u>0$ or in other words, if $a>1$, as in the OP,
then there is just one solution
\begin{align}
z&=
\exp\left(\W\left(\ln(a)\exp(\ln(a))\right)\right)
=\exp(\ln(a))=a
\tag{7}\label{7}
.
\end{align}
But if $u<0$ or in other words, if $0<a<1$
then there are two real solutions:
\begin{align}
z_0&=
\exp\left(\Wp\left(\ln(a)\exp(\ln(a))\right)\right)
\tag{8}\label{8}
,\\
z_1&=
\exp\left(\Wm\left(\ln(a)\exp(\ln(a))\right)\right)
\tag{9}\label{9}
.
\end{align}
One of the solutions would always be $a$:
if $-1<\ln(a)<0$ then $z_0=a$,
else if $\ln(a)<-1$ then $z_1=a$.
For example, if $a=\tfrac1{10}$, then
\begin{align}
z_0&=
\exp\left(\Wp\left(\ln(\tfrac1{10})\exp(\ln(\tfrac1{10}))\right)\right)
\approx 0.7292411
\tag{10}\label{10}
,\\
z_1&=
\exp\left(\Wm\left(\ln(\tfrac1{10})\exp(\ln(\tfrac1{10}))\right)\right)
=
\exp\left(\ln(\tfrac1{10})\right)
=\tfrac1{10}
\tag{11}\label{11}
.
\end{align}
$\endgroup$
