Solve the equation $x^{x^5} = 5$ over $\Bbb R.$ Find real numbers $x$ (if any) such that $x^{x^5} = 5.$ 
I have shown that $x \notin \Bbb Z.$ Does there exist any $x \in \Bbb R \setminus \Bbb Z$ which satisfies the above equation? Any help will be highly appreciated.
 A: \begin{align} 
x^{x^5}&=5
\tag{1}\label{1}
\end{align}
Equation \eqref{1} can be solved formally
with the help of the Lambert W function:
\begin{align} 
x^5\ln x&=\ln 5
,\\
5x^5\ln x&=5\ln 5
,\\
x^5\ln (x^5)&=5\ln 5
,\\
\ln (x^5)\,\exp(\ln (x^5))&=\ln 5\exp(\ln5)
.
\end{align}
At this point we can apply the Lambert W function:
\begin{align} 
\operatorname{W}(\ln (x^5)\,\exp(\ln (x^5)))
&=
\operatorname{W}(\ln 5\exp(\ln5))
,
\end{align}
and since the argument 
$\ln 5\exp(\ln5)>0$,
there is just one solution
\begin{align} 
\ln (x^5)&=\ln5
,\\
x^5&=5
,\\
x&=5^{\tfrac15}\approx 1.379729661
.
\end{align} 

Edit
Or, less formally, without Lambert W function:
\begin{align} 
(x^{x^5})^5&=5^5
,\\
x^{5x^5}&=5^5
,\\
(x^5)^{(x^5)}&=(5)^{(5)}
,\\
x^5&=5
.
\end{align}
A: The hint.
$\sqrt[5]5$ is the root. 
Now, prove that it's an unique root for which consider two cases:


*

*$0<x\leq1$;

*$x>1$.
A: As $x^{x^5} = 5$, we can always substitute the $5$ on the index with $x^{x^5}$
$$\begin{array}{rcll}
x^{x^5} &=& 5\\
x^{x^{x^{x^5}}} &=& 5\\
\end{array}$$
Repeat infinitely many times and you will get:
$$x^{x^{x^{x^{\cdots}}}} = 5$$
The index that $x$ is raised to, is same as the entire "power tower", which is 5:
$$x^{\boxed{x^{x^{x^{\cdots}}}}} = 5$$
$$x^5 = 5$$
We get $x = \sqrt[5]{5}$.
A: Here are graphs that show the solution, $x = \sqrt[5]{5}$:

