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.)
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.