I’m having trouble following the equation below: $$\begin{align} E_{period} & = \int_0^{T_0} \lvert e^{j\omega_0t} \rvert \,dt \\ & = \int_0^{T_0} 1 \cdot dt = T_0 \end{align} $$
I know that $ \cos^2(x) + \sin^2(x) = 1 $, so I'm assuming they are somehow using that identity, but I can't figure out how they get there.
$$e^{j\omega_0t}= \cos{\omega_0t}+j\sin{\omega_0t}$$ so
$$\left(e^{j\omega_0t}\right)^2 = \cos^2(\omega_0t) + 2j\cos(\omega_0t)\sin(\omega_0t) - \sin^2(\omega_0t)$$
This is as far as I can get. I know $2\cos(x)\sin(x) = \sin(2x)$, but I don't think that helps me here. Really stumped, any help is much appreciated.
Hint: $$\vert\exp(jwt)\vert^2 = \exp(jwt) \cdot \exp(-jwt).$$