We can define ${}^nx$ as $\underbrace{\displaystyle {x^{x^{\cdot ^{\cdot ^{x}}}}}}_{n\text{ times}}$ (Tetration). I conjecture that ${}^ni$ is complex for all $n \ge 3, n \in \mathbb{N}$.
I've attempted to prove this via induction and the exponential form of of a complex number ($i = e^{i\pi/2}$), and believe that I succeeded. However, my proof is rather shaky, and I'm wondering about a more rigorous proof? So, how would you prove
${}^ni \in \mathbb{C}, \forall n \ge 3, n \in \mathbb{N}$
I won't include my full proof as it's rather long, but a quick summary is something like this:
We'll use proof by induction, starting with $n = 3$
Let $n = 3$. Therefore
$$ \begin{align} {}^3i & = i^{i^i} \\ & = i^{e^{-\pi/2}} \\ & = (e^{i\pi/2})^{e^{-\pi/2}} \\ & = e^{(i\pi e^{-\pi/2})/2} \\ & = \cos(\frac{\pi}{2} e^{-\pi/2}) + i\sin(\frac{\pi}{2} e^{-\pi/2}) \\ & \in \mathbb{C} \end{align} $$
Now, assume that the statement is true for $n = k$, i.e.
$$ {}^ki = \cos\theta + i\sin\theta, \: \theta \ne m\pi $$
Let $n = k+1$. Therefore
$$ \begin{align} {}^ni & = {}^{k+1}i \\ & = i^{({}^ki)} \\ & = i^{\cos\theta + i\sin\theta} \\ & = i^{\cos\theta}i^{i\sin\theta} \\ & = (e^{i\pi/2})^{\cos\theta}(i^i)^{\sin\theta} \\ & = (e^{(i\pi\cos\theta)/2})(e^{-\pi/2})^{\sin\theta} \\ & = r\left(\cos\left(\frac{\pi}{2}\cos\theta\right) + i\sin\left(\frac{\pi}{2}\cos\theta\right)\right) \\ & \in \mathbb{C} \end{align} $$
where $r = e^{-\pi/2}$. Now, I thought that the only way this may not be complex is if $\sin\left(\frac{\pi}{2}\cos\theta\right) \ne 0$, so I performed proof by induction a second time on $\theta$, which proved that $\sin\left(\frac{\pi}{2}\cos\theta\right) \ne 0$ for the relevant $\theta$.
In addition, I've verified it using this Jelly program up to $n = 1000$
As a side note, is this a "known" fact/proof?