Find all values of $(1+i)^{1+i}$ $$z=(1+i)^{1+i}$$
I'm having trouble with this one. I got as far as the below, but then I got stuck. Could someone give me a hint??
$$\ln(z)=(1+i)\ln(1+i)$$

After reading the hints + suggested answer:
$$z=(1+i)^{1+i}$$
$$z=e^{(1+i)ln(1+i)}$$
$$z=e^{(1+i)ln|1+i|+(arg(1+i)+2\pi k)i}$$
$$z=e^{ln(\sqrt{2})+\pi i/4+2\pi ik+iln(\sqrt{2})-\pi /4-2\pi k}$$
Is it ok to leave my final answer like this?

$\forall n\in \mathbb{Z}$,
\begin{align*}
  e^{2n\pi i} &= 1 \\
  1+i &=
  \exp \left[ \frac{\ln 2}{2}+i\pi \left( 2n+\frac{1}{4} \right) \right] \\
  (1+i)^{1+i} &=
  \exp \left[
         \frac{1+i}{2} \ln 2+i(1+i) \left( 2n+\frac{1}{4} \right) \pi
       \right] \\
  &= 
  \exp \left[
          \frac{\ln 2}{2}-\left( 2n+\frac{1}{4} \right) \pi+
         i\left( \frac{\ln 2}{2}+2n\pi+\frac{\pi}{4} \right)
       \right] \\
  &= \sqrt{2} \, \exp \left[
                        -\left( 2n+\frac{1}{4} \right) \pi+
                        i\left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)
                      \right] \\
  &= \frac{\sqrt{2}}{e^{\left( 2n+\frac{1}{4} \right) \pi}}
     \left[
        \cos \left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)+
       i\sin \left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)
     \right]
\end{align*}
