Let us take a look at this equation.
$$e^{e^x}=1$$
when $e$ is the base of the natural logarithm.
It may seem intuitive that there are no solutions in any field. After all, $e^0=1$, so there can not be anything $x$ such that $e^x=0$.
Is this true? Are there absolutely no solutions?
Let us take Euler's identity, where $i^2=-1$
$$e^{i\pi}+1=0$$
Subtracting $1$ from both sides, we get
$$e^{i}=-1$$
What happens when we square both sides?
$$e^{i2}=1$$
This means that i2 is also a natural logarithm of $1$.
$$e^x=i2$$
Taking the natural log of both sides:
$$x=\ln(i2)=\ln(2)+\ln()+\ln(i)=\ln(2)+\ln()+i/2$$
This is about $1.8379+i1.5708$. It is not yet known if this is transcendental. However, if it is algebraic, then $\ln(2)+\ln()$ (about $1.8379$) must be irrational (and transcendental!).
But wait.
$$(e^{i2})^2=1^2=e^{i4}=1$$
$$(e^{i2})^3=1^3=e^{i6}=1$$
If $z$ is an integer
$$(e^{i2})^z=1^z=e^{i2z}$$
All $i2z$ are natural logarithms of $1$ for all $z$. So the original equation above has the solution set:
$$x=\ln(2z)+\ln()+i/2$$
For all nonzero $z$.
Now let $e=10$. There should be a countable infinitude of solutions.