Complex solutions of $e^{e^z}=1$ I know that the solution of $e^z = 1$ is $2\pi k$, for integers $k.$  Note $z$ is complex.  I just don't really know how to start tackling $e^{e^z} = 1$

It seems that this means we want $e^z = 2\pi k$, i.e $z = \log(2\pi k)$, but this somehow seems like it's too simple to be right.
First off, there is an error in your question: if $\omega \in \mathbb{C}$, then
$$ \mathrm{e}^{\omega} = 1 \implies \omega = 2\pi k\color{red}{i}, $$
where $k$ is any integer.  Note that this is a purely imaginary number, rather than the real value $2\pi k$ in your question.  Taking $\omega = \mathrm{e}^{z}$, we have
$$ \mathrm{e}^{\mathrm{e}^z} = 1
\implies \mathrm{e}^z = 2\pi k i, $$
where $k$ is any integer.

Next, it might be helpful to (rather pedantically) expand out the right-hand side of the equation in terms of exponentials:
$$ \mathrm{e}^z
= 2\pi k i
= (2\pi k) (i)
= \mathrm{e}^{\log(2\pi k)} \mathrm{e}^{i\left(\frac{\pi}{2} + 2\pi\ell\right)}
= \mathrm{e}^{\log(2\pi k) + i\left(\frac{\pi}{2} + 2\pi\ell\right)},
$$
where $\ell$ is an arbitrary integer and $\log$ is the the natural logarithm defined on $\mathbb{R}$; we are using Euler's formula to write out $i$ as an exponential expression.  Now, equating the real and imaginary parts on the left- and right-hand sides of the equation, we have
$$ \Re(z) = \log(2\pi k)
\qquad\text{and}\qquad
\Im(z) = \frac{\pi}{2} + 2\pi\ell, $$
where $k \in \mathbb{Z}\setminus\{0\}$ (note that the logarithm of zero is undefined, hence $k$ cannot be zero) and $\ell \in \mathbb{Z}$.  In other words, with $k$ and $\ell$ as above, we conclude that
$$ z = \log(2\pi k) + i \left( \frac{\pi}{2} + 2\pi\ell \right). $$
The solution should be for $k\neq 0$

$$e^z=2ik\pi\implies z=2ik_1\pi+\log 2ik\pi=2ik_1\pi+\log (2k\pi\cdot e^{\frac{\pi}2i+2ik\pi})=\\=2ik_1\pi+\log 2k\pi+\frac{\pi}2i+2ik\pi=\\=\log 2k\pi+\frac{\pi}2i+2in\pi$$

Refer here for Complex logarithm.