I want to prove or disprove that the product of two transcendental is transcendental: (However, not using an inverse identity such as $\pi$ and $\frac1\pi$)
My attempt:
Proof using Hilbert's number: $2^\sqrt2$
The product of Hilbert's number I contend is transcendental: $(2^\sqrt2)^2$ =$2^\sqrt2 * 2^\sqrt2 = 2^{\sqrt2+\sqrt2}= 2^{irrationalnumber}$
$\sqrt2+\sqrt2$ is irrational because there sum has no parts that cancel out.
Thus, $2^{irrationalnumber}$ by Gelfond-Schneider Thorem, any number of the form $a^b$ is transcendental where $a$ and $b$ are algebraic and $b$ is not a rational number.
I dislike the $2^{irrationalnumber}$ and the fact that I used the same two transcendental is the a more stronger proof that use two distinct and no inverse like the questions asks?
The statement "the product of two transcendentals is transcendental" means "the product of any two transcendentals is always transcendental". It is true that $2^{\sqrt{2}} \times 2^\sqrt{2} = 2^{2 \sqrt{2}}$ is transcendental, but this is just an example to show that the product of two transcendentals can be transcendental.
If you don't like $\pi$ and $1/\pi$, take any nonzero algebraic number $c$ and any transcendental number $a$. Then $b = c/a$ will also be transcendental, but $ab = c$ is not transcendental.