Why is $\left|e^{iat} e^{-st}\right| \le \left| e^{iat}\right| \left| e^{-st}\right| \le e^{-st}$ true? My text states:


  $\left|e^{iat} e^{-st}\right| \le \left| e^{iat}\right| \left| e^{-st}\right| \le e^{-st}$
  
  as $\left| e^{iat} \right|=1$
  where $a,s,t \in \mathbb{R}$


I thought the last part of the inequality should read:


  $\left| e^{iat}\right| \left| e^{-st}\right| = e^{-st}$


since $\left| e^{-st}\right|$ is always positive.

I am also not sure how the middle expression of the inequality is arrived at. I do know that:
$$ \begin{align} a,b \in \mathbb{R} \Rightarrow &amp; \left|ab\right| = \left| a \right| \left|b\right|\end{align}$$
By the property of the absolute value function.

But in this case, the left hand side of the inequality is the modulus of a complex number so I am not sure if the above rule still applies.I'm also not able to come up with a geometric argument to justify why the inequality must be true.

Please help me see why the given inequality must be true especially why it has to be a weak inequality instead of an equality.
I think it's equal, since complex norm is also multiplicative!
I think it's equal, since complex norm is also multiplicative!