Does this property hold true for natural number values of $k$?

While I was playing around with $i$, I discovered an interesting pattern where $i^{2^{-k}}=i^{10^{-k}}$ where $k∈ℕ$.

Here were my steps:
\begin{align}
i^{2^{-k}} & = i^{\frac{1}{2^{k}}} \\
& = i^{\frac{1}{2^{k}}\cdot\frac{5^{k}}{5^{k}}} \\
& = i^{\frac{5^{k}}{10^{k}}} \\
& = i^{10^{-k}} \\

\end{align}

The reasoning behind why I was able to ignore the $5^{k}$ in the numerator is because $i^{5^{k}}$ where $k∈ℕ$ will always be equal to $i$.

I was hoping if someone else could check if I am correct or incorrect for assuming this for all natural number values of $k$.
(I would also like to add I am currently a highschool student that is just curious about math and that curiousity has enabled to come up with different conjectures just for fun.)

Answer

Your conjecture is false:
$$
i^{1/2}\ne i^{1/10}.

$$

The reason for the failure is the fact that the relation $(z^a)^b=z^{ab}$ does not hold for complex numbers.