Determine all the values of $1^{\sqrt{2}}$

Hint: Recall that $e^{x+iy}$ where $x,y\in\mathbb{R}$ is equal to $e^x(\cos(y)+i\sin(y))$. What complex values $z$ give $e^z=1$?

Expanding the hint: We start with $e^{\sqrt{2}\log(1)}$. Solving the above equation, we see that $e^{z}$ is one at precisely $2\pi ik$ for $k\in\mathbb{Z}$, so $\log(1)=\{2\pi i k|k\in\mathbb{Z}\}$. Plugging in, we have the set $e^{2\sqrt{2}\pi i k}$ for $k\in\mathbb{Z}$. Expanding according to the above, we have $e^{2\sqrt{2}\pi i k}=\cos(2\sqrt{2}\pi k)+i\sin(2\sqrt{2}\pi k)$ and we arrive at the answer.

The punchline here is that $\log(z)$ is still the inverse of $e^z$, but $e^z$ is no longer 1-1 and therefore $\log(z)$ can only be a local inverse, and there's some choice involved in which branch to pick.