Does limlimits_(n→∞)ni^n exist?
Does $\ \lim\limits_{n\to\infty}ni^n$ exist?
Explanation of Method
Split by $n\bmod4$ (the period of $i^n$) to see that the sequence heads to infinity along four different rays; conclude no limit exists in $\mathbb C$, but a limit of $\infty$ does exist in the extended plane $\overline{\mathbb C}$.
Step 1 · The four residue classes
$$ i^n=\begin{cases} i & n\equiv1\!\!\pmod4\\ -1 & n\equiv2\!\!\pmod4\\ -i & n\equiv3\!\!\pmod4\\ 1 & n\equiv0\!\!\pmod4\end{cases} \qquad\Longrightarrow\qquad ni^n=\begin{cases} ni\\ -n\\ -ni\\ n\end{cases} $$
Step 2 · Behavior of each subsequence
All four subsequences are unbounded: $ni\to\infty$ along the positive imaginary axis, $-n\to\infty$ along the negative real axis (i.e. $-\infty$), $-ni\to\infty$ along the negative imaginary axis, and $n\to\infty$ along the positive real axis — four genuinely different directions.
Step 3 · In $\mathbb C$: no limit
Since the sequence is unbounded and (were it to converge) a limit would have to be a single finite complex number that every subsequence approaches, but these four subsequences pull toward four incompatible directions, $\lim_{n\to\infty}ni^n$ does not exist as a finite complex number.
Step 4 · In $\overline{\mathbb C}$: the limit is $\infty$
Every term satisfies $|ni^n|=n|i^n|=n\to\infty$, regardless of direction. Since $\overline{\mathbb C}=\mathbb C\cup\{\infty\}$ has only one point at infinity (independent of direction, by the Riemann-sphere topology), this uniform blow-up of $|z_n|$ means:
$$ \boxed{\lim_{n\to\infty}ni^n=\infty \ \text{ in } \overline{\mathbb C}} $$
Verification
$$ z_1=i,\quad z_2=-2,\quad z_3=-3i,\quad z_4=4,\quad z_5=5i,\quad z_6=-6,\ldots $$
the moduli $1,2,3,4,5,6,\ldots$ grow without bound while the direction cycles through the four axis directions.
Result
A clean illustration of why the extended complex plane is useful: it "resolves" divergences to infinity into a single well-defined limit, even when the direction of divergence keeps changing.