Solving $e^{e^z}=1$: am I missing something? I solved the equation $e^{e^z}=1$ and it seemed to easy so I suspect I must be missing something.

Would someone please check my answer?

My original answer:
$e^{e^z}=1$ if and only if $e^z = 2\pi i k$ for $k\in \mathbb Z$ if and only if $z=\ln(2\pi i k)$ for $k\in \mathbb Z$.
Edit
After reading the comments and answers I tried to do it again. Unfortunately, I still do not get the same result as in the answers.
My second attempt:
We have 
$$ e^x = 1 \iff x = 2 \pi i k$$
hence 
$$ e^z = 2 \pi i k$$
for some $k$ in $\mathbb Z$. 
Letting $e^z = e^x (\cos y + i \sin y)$ we get 
$$ e^x \cos y + i e^x \sin y = 2 \pi k i$$
which implies that $\cos y = 0$ which happens if and only if $y_j = {\pi \over 2} + \pi j$ where $j\in \mathbb Z$. At $y_j$ we have
$\sin y = \pm 1$ hence if $j$ is even
$$ e^x = 2 \pi i k$$
and if $j$ is odd 
$$ e^x = -2 \pi i k$$
Hence if $j$ is even,
$$ x = {\pi \over 2} + \ln(2 \pi k)$$
and if $j$ is odd,
$$ x = {3\pi \over 2} + \ln(2 \pi k)$$
So we see that the solutions are
$$
z_{t,k}=\begin{cases}
 {\pi \over 2} + \ln(2 \pi k) + i ({\pi \over 2} + 2t \pi )\\ 
 {3\pi \over 2} + \ln(2 \pi k) + i({\pi \over 2} + (2 +1)t \pi )
\end{cases} 
$$
for $k,t \in \mathbb Z$.

What am I doing wrong?

 A: Since $1$ can be written $1=e^0$ it follows that a first solution is
$$e^z=i2\pi k,\qquad (k \in \mathbb{Z})$$
If $k=0$ there are no solutions since $e^z$ is never zero.
If $k>0$ write $i2 \pi k$ in exponential polar form and you should find that
$$|i2 \pi k|=2 \pi |k| = 2 \pi k, \qquad (\text{since $k>0$})$$
So
\begin{align}
i 2 \pi k &= e^{i \pi /2}e^{\ln(2 \pi k)} \\
          &= e^{i \pi /2 + \ln(2 \pi k)}
\end{align}
Hence solution for $k>0$ is of the form
$$z=\ln(2 \pi k)+i\left(\frac{\pi}{2}+2 \pi n \right), \qquad \text{for $n \in \mathbb{Z}, k \in \mathbb{Z} \cap (0, + \infty)$}$$
I leave the case $k < 0$ as an exercise for the OP.
A: The problem is that your expression "$\ln 2\pi ik$" is multivalued (worse, it is undefined if $k=0$).
One may write $$e^{e^z}=1 \iff e^z = 2\pi i k \stackrel{k\neq 0}{\iff} z =
\begin{cases}
\ln 2\pi k + \frac{\pi i}{2}(4n+1), & k>0\\
\ln -2\pi k + \frac{\pi i}{2}(4n-1), & k<0
\end{cases}$$
So the solutions are $$z_{n,k} = \ln2\pi |k| + \frac{\pi i}{2}\left(4n+\frac{k}{|k|}\right)$$ for integral $n$ and nonzero integral $k$, and "$\ln$" is the real-valued function of a positive real variable (note that since $k$ is nonzero, one has that $\frac{k}{|k|} = \pm 1$ according to the sign of $k$).
A: I believe you are making too many jumps
Consider that in Approach 1 you said


*

*$e^{e^z}=1$ if and only if $e^z = 2\pi i k$


*$e^z = 2\pi i k$ for $k\in \mathbb Z$ if and only if $z=\ln(2\pi i k)$ for $k\in \mathbb Z$

and in Approach 2, you said

$$ e^x = 1 \iff x = 2 \pi i k$$
hence
$$ e^z = 2 \pi i k$$


Of course we see now that (2) is wrong. While (1) is right (Or not?), (1) seems like cheating death (or cheating vanishment-in-a-puff-of-logic). You're assuming $e^{iy}$ and $e^z$ behave too similarly, if not outright the same, to $e^x$. I suggest you expand $e^{e^z}$ by definition instead of making jumps as to which term should equal some multiple of $\pi$.
Let's start with the $z$:
$$e^{e^z}= e^{e^{x+iy}}$$
and the inner $e^z = e^{x+iy}$:
$$e^{e^{x+iy}}=e^{e^{x}e^{iy}}$$
Now for the $e^{iy}$:
$$e^{e^{x}e^{iy}}=  e^{e^{x}(\cos(y)+i\sin(y))}=e^{e^{x}\cos(y)+ie^{x}\sin(y)}$$
Let's proceed to the outer $e^{f(x,y)+ig(x,y)}$:
$$e^{e^{x}\cos(y)+ie^{x}\sin(y)}=e^{e^{x}\cos(y)}e^{ie^{x}\sin(y)} $$
Now for the $e^{ig(x,y)}$:
$$e^{e^{x}\cos(y)}e^{ie^{x}\sin(y)}= e^{e^{x}\cos(y)}(\cos(e^{x}\sin(y))+i\sin(e^{x}\sin(y)))$$
Finally, we can bring in the 1:
$$e^{e^{x}\cos(y)}(\cos(e^{x}\sin(y))+i\sin(e^{x}\sin(y))) = e^{e^{x}\cos(y)}\cos(e^{x}\sin(y))+ie^{e^{x}\cos(y)}\sin(e^{x}\sin(y)) = 1+0i$$
$$\iff e^{e^{x}\cos(y)}\cos(e^{x}\sin(y)) = 1, e^{e^{x}\cos(y)}\sin(e^{x}\sin(y)) = 0$$
$\therefore,$ we reduce a complex equation into a system of 2 real equations, whose solutions are:
$$(x,y) = (\ln(2m \pi),\frac{\pi}{2} + 2 l \pi)$$
This can be shown to be equivalent to WA's answer:
$$z = 2 \pi i n_2 + \ln(2 i \pi n_1)$$
