Are those two numbers transcendental? Suppose, $u$ solves the equation $$u^u=\pi$$ and $v$ solves the equation $$v\cdot e^v=\pi$$
So, we have $u=e^{W(\ln(\pi))}$ and $v=W(\pi)$. $u$ and $v$ should be the real solutions (in this case, they are unique). If someone can prove that $u$ and $v$ are transcendental, the next step would be to show that every solution is transcendental, but for the moment, the real case is sufficient.


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*Are $u$ and $v$ transcendental ?


The Lindemann-Weierstrass-theorem does not help here because $\pi$ is transcendental and $\ln(\pi)$ and $W(\ln(\pi))$ are probably (does anyone know a proof ?) transcendental.
The algdep-command of PARI/GP does neither indicate the algebraicy of $u$ nor the algebraicy of $v$. So, if $u$ or $v$ is algebraic, the minimal polynomial must have high degree or coefficients with a large absolute value.
 A: $$u^u=ve^v=\pi\Rightarrow u^u=e^{v+\ln v}=\pi\Rightarrow\begin{cases}u\approx1.85411\\v\approx 1.07364\end{cases}$$
The function $f(x)=x^x$ is injective for $x\gt 1$ and so is $g(x)=xe^x$ for $x\gt 0$ hence the finded $u$ and $v$ are unique real solutions.
Besides $u$ cannot be rational because if
$$x=\left(\frac ab\right)^{\frac ab}\iff b^ax^b-a^a=0$$ then $x$ is algebraic but we know that $\pi$ is trascendental. It follows that $u$ is algebraic irrational or trascendental.
On the other hand, Geldfond-Schneider theorem ensures that if $a$ and $b$ are algebraic with $a\notin\{0,1\}$ and $b$ irrational then $a^b$ is trascendental so it is not impossible that $u$ would be algebraic but it could be trascendental also. At the current state of knowledge on this topic, we can not guarantee something more on the nature of $u$ besides of it is not rational. We only know that $u$ is algebraic of degree greater than $1$ or trascendental, nothing more. 
Instead, $v$ could be rational provided $e^v$ is trascendental and we  ran into the same impasse. I stop here.
