Simplifying frequency response equation
$$1 - e^{-4j\omega} = e^{-2j\omega}(e^{2j\omega}  - e^{-2j\omega}) \tag{1}$$

Now,

$$ \sin(2\omega) = \frac{e^{2j\omega}  - e^{-2j\omega}}{2j} \tag{2}$$

Equation 2 is a consequence of Euler's formula. Multiply and divide by $2j$ in (1) and use the identity (2) in equation 1 we have:

$$1 - e^{-4j\omega} = 2je^{-2j\omega}\sin(2\omega) \tag{3}$$

Now $j = e^{\frac{j\pi}{2}}$, substituting in 3 we have 

$$1 - e^{-4j\omega} = 2e^{-2j\omega}e^{\frac{j\pi}{2}}\sin(2\omega) \tag{4}$$

The term $e^{-2j\omega}$ in (1) is represntative of the delay of the envelope in time domain, the term $(e^{2j\omega}  - e^{-2j\omega}) $ is reprsentative of the magnitude response and actual shape of the filter in frequency domain.

A general trick in simplifying expressions of FIR responses type is to tap expressions around the mid point of the impulse response taps, you could use this strategy to pair terms in responses with more than 2 terms as well