I've written the below 2 claims.
Claim1.
$$ \left| \prod_{i=1}^{n} a_{i} \right|=\prod_{i=1}^{n} \left|a_{i} \right|$$
$$ \left\{ a \right\} :=\text{sequence which holds} ~a_{i} \in\mathbb{R} $$
$$n:=n(a)$$
Proof.
$$ \left\{ b \right\} :=\text{sequence which holds}~ \left| a_{i} \right| \cdot (-1)^{b_{i}} = a_{i}~~~ \text{by} ~~~b_{i} \in\left\{0,1\right\} $$
$$ \left| \prod_{i=1}^{n} a_{i} \right|= \left|\prod_{ i=1 }^{ n } \left| a_{i} \right| \left( -1 \right) ^{b_{i}} \right| $$
$$=\left|\left(-1\right)^{\text{0 or 1}}\prod_{ i=1 }^{ n } \left| a_{i} \right| \right| $$
$$=\left|\prod_{ i=1 }^{ n } \left| a_{i} \right| \right| $$
$$=\prod_{ i=1 }^{ n } \left| a_{i} \right|$$
$$ \left| \prod_{i=1}^{n} a_{i} \right|=\prod_{i=1}^{n} \left|a_{i} \right|$$
$$ \text{QED.} $$
Claim2.
$$ a,b,c \in\mathbb{R} $$
$$j:=\text{imaginary number}$$
$$ \left| a+b \exp\left(- j\omega \right) + c \exp\left(- 2j\omega \right) \right| = \left| a+b \exp\left( j\omega \right) + c \exp\left( 2j\omega \right) \right| $$
Proof.
$$ \left| a+b \exp\left(- j\omega \right) + c \exp\left(- 2j\omega \right) \right| = \left| a+b \exp\left( j\omega \right) + c \exp\left( 2j\omega \right) \right| $$
$$ \left| a\exp\left(j \cdot 0\right) +b \exp\left(- j\omega \right) + c \exp\left(- 2j\omega \right) \right| = \left| a\exp\left(j \cdot 0\right)+b \exp\left( j\omega \right) + c \exp\left( 2j\omega \right) \right| $$
I assumed the each absolute value inside the absolute value bars as sum of each magnitude of vector of 2d cartesian system( or polar coordinates syetem). So the equation holds.
I've drawn the below rough image.
However if we develop the exponential functions,
$$ \left| a+b \exp\left(- j\omega \right) + c \exp\left(- 2j\omega \right) \right| = \left| a+b \exp\left( j\omega \right) + c \exp\left( 2j\omega \right) \right| $$
$$ \left| a+b \left( \cos\left(-\omega \right) +j\sin\left(-\omega \right) \right) + c \left( \cos\left(-2 \omega \right) +j\sin\left(-2 \omega \right) \right) \right| = \left| a+b \left( \cos\left(\omega \right) +j\sin\left(\omega \right) \right) + c \left( \cos\left(2 \omega \right) +j\sin\left(2 \omega \right) \right) \right| $$
$$ \left| a+b \left( \cos\left(\omega \right) -j\sin\left(\omega \right) \right) + c \left( \cos\left(2 \omega \right) -j\sin\left(2 \omega \right) \right) \right| = \left| a+b \left( \cos\left(\omega \right) +j\sin\left(\omega \right) \right) + c \left( \cos\left(2 \omega \right) +j\sin\left(2 \omega \right) \right) \right| $$
$$ \left| a+b \cos\left(\omega \right) -bj\sin\left(\omega \right) + c \cos\left(2 \omega \right) -cj\sin\left(2 \omega \right) \right| = \left| a+b \cos\left(\omega \right) +bj\sin\left(\omega \right) + c \cos\left(2 \omega \right) +cj\sin\left(2 \omega \right) \right| $$
So ,of above equation,$ ~~a ~,~ b \cos\left(\omega \right) ~,~ c \cos\left(2 \omega \right) $ are shared in each formula however the signs of $ b j\sin\left(\omega\right) ~,~ c j\sin\left( \omega \right) $ are opposite.
So seemingly I think that the equation cannot be held.
What I've been missing?
For Claim 2, you have a small typo from mis-applying Euler's formula. You should end up with
$$|a + b \cos \omega - ib \sin \omega + c \cos (2\omega) - ic \sin (2\omega)| = |a + b \cos \omega + ib \sin \omega + c \cos (2\omega) + ic \sin (2\omega)|$$
Then using the fact that $|p+jq|^2 = p^2 + q^2$ when $p,q \in \mathbb{R}$, you can check that the square of both sides of the above equation are $$(a+b \cos \omega + c \cos (2\omega))^2 + (b \sin \omega + c \sin (2\omega))^2.$$