Is there an exact value to the minimum of the infinite tetration of $x$? 
Is there an exact value to the minimum of the function $$f_k(x)=\underbrace{x^{x^{x^{.^{.^.}}}}}_{2k\,\text{times}}$$ as $k\to\infty$, where $k=1,2,3,\cdots$?

This visualisation in Desmos shows that the minimum is around $0.3792$. 
I have specifically stated $2k$ as the function is strictly increasing for odd tetrations.
I am certain that it will not tend to $0$ since for every $x^x$ I add to the tetration, it decreases by $0.0004$, then $0.0003$ and $0.0002$. 
Differentiation looks extremely tedious to do...
So is there a rigorous approach to find the exact value, if there exists one, of $\min(f_\infty(x))$?
If not, is there a method to find a lower bound?
 A: 
Claim: The minimum $1/e \approx 0.36788$ is achieved at $x = e^{-e} \approx 0.065988$.

We proceed by first establishing that the minimum is upper bounded by $1/e$ and then show that it is also lower bounded by $1/e$.
Proof of upper bound. Let $y_{0}=1$ and define the iteration
$$
y_{n+1}=x^{y_{n}}.
$$
Note that
$$
y_{1}=x,\qquad y_{2}=x^{x},\qquad\text{etc.}
$$
tends to the infinite tetration. It is known that this iteration converges (possibly to $\infty$) for all $x\geq e^{-e}$. Define $y_{\infty}$ as the function which associates to each $x \geq e^{-e}$ the limit $\lim_{n\rightarrow\infty}y_{n}$. Note that if $y_\infty(x)$ is finite, then
$$
y_{\infty}(x)=\lim_{n\rightarrow\infty}y_{n+1}=\lim_{n\rightarrow\infty}x^{y_{n}}=y^{y_{\infty}(x)}
$$
and hence
$$
(y_{\infty}(x))^{1/y_{\infty}(x)}=x.
$$
Solving for $y_{\infty}(x)$,
$$
y_{\infty}(x)=\frac{W(\log(1/x))}{\log(1/x)}.
$$
Let $x_0=1$ and define also the iteration
$$
x_{n+1}=x^{x^{x_{n}}}.
$$
Since it is monotone, this iteration converges (possibly to $\infty$) for all $x\geq0$. Define $x_{\infty}$ as the function which associates to each $x \geq 0$ the limit $\lim_{n\rightarrow\infty}x_{n}$. Since $x_{\infty}=y_{\infty}$ on $[e^{-e},\infty)$,
$$
\min_{x\geq0}x_{\infty}(x)\leq y_{\infty}(e^{-e})=1/e.
$$
Proof of lower bound. Fix $x$ and let $a=x_{\infty}(x)$ for brevity. Then,
$$
x^{x^{a}}=a\implies x=a^{1/(x^{a})}\implies x^{a}=a^{a/(x^{a})}\implies(x^{a})^{x^{a}}=a^{a}.
$$
Therefore,
$$
a^{a}\geq\min_{u\geq0}u^{u}=e^{-1/e},
$$
which in turn implies
$
a\geq1/e
$.
