$$\begin{align} e^{i\theta}&=\cos\theta+i\sin\theta \\[0.5em] \left|e^{i\theta}\right|&=\sqrt{\cos^2\theta+\sin^2\theta} \end{align}$$
I get that the magnitude of $e^{i\theta}$ has to be $1$, but why does that $i$ squared become $1$? Isn't $i$ squared supposed to be $-1$?
On
In the real numbers, $|z| = \sqrt{z^2}$. However, in the complex numbers this is no longer true. Instead, we say that $|z| = \sqrt{z\bar{z}}$ where $\bar{z}$ is the complex conjugate of $z$, with $\overline{x + iy} = x - iy$. Notice that $(x + iy)(x - iy) = x^2 - i^2 y^2 = x^2 + y^2$, which guarantees that $|z|$ is always a non-negative real number. Also, notice that if $z$ is real, then $\bar{z} = z$ so that $|z| = \sqrt{z^2}$ still.
Then if $z = \cos \theta + i \sin \theta$, $|z| = \sqrt{(\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta)} = \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt{1} = 1$.
Think about a right triangle with height $\sin \theta$ and base $\cos \theta$, the number $\cos\theta + i\sin\theta$ is the tip of the triangle in the complex plane. So to compute the hypothenuse you do the pythagorean theorem which is the formula for $\left|e^{i\theta}\right|$ you mentioned.