Product of two transcendental numbers is transcendental Let $\alpha,\beta$ be transcendental numbers. Which of the followings are true?
1)$\alpha\beta\ \text{ is transcendental}$.
2)$\mathbb{Q}(\alpha)\ \text{is isomorphic to }\mathbb{Q}(\beta)$
3)$\alpha^\beta\ \text{is transcendental }$
4)$\alpha^2\ \text{is transcendental}$
I know option 4 is true. And I feel option 1 is also true, but I don't know the exact  reason. While I have no idea  for the remaining two. 
 A: *

*Let $\alpha$ be any transcendental. Then $\beta := \frac{1}{\alpha}$ is transcendental and $\alpha\cdot \beta =1$. Thus 1. is false.

*Consider $f \colon \mathbb Q(\alpha) \to \mathbb Q(\beta), \frac{x_0 + x_1 \alpha + \ldots + x_n \alpha^n}{y_0 + y_1 \alpha + \ldots + y_m \alpha^m} \mapsto \frac{x_0 + x_1 \beta + \ldots + x_n \beta^n}{y_0 + y_1 \beta + \ldots + y_m \beta^m}$, where the $x_i,y_j$ are rationals. Verify that $f$ is indeed an isomorphism.

*Notice that $e$ and $\ln 2$ are transcendental, but $e^{\ln 2} = 2$. So this is false.

*If $\alpha$ is transcendental, then $\alpha^n$ is transcendental for every $n \in \mathbb N$ (if there were a nontrivial polynomial $p(x) \in \mathbb Z[x]$ with $p(\alpha^n) = 0$, then let $q(x)$ be the polynomial obtained from $p(x)$ by replacing every $x^k$ in $p(x)$ with $x^{n\cdot k}$. Then $q(x)$ is nontrivial and $q(\alpha) = p(\alpha^n) = 0$, so $\alpha$ is algebraic - Contradiction!)

A: Here is a proof that point 3 is false that doesn't use any difficult theorems.
As $\gamma$ ranges through all transcendental numbers, the number $2^\gamma$ takes on uncountably many values. Since only countably many of these values can be algebraic, there must exist a transcendental $\gamma$ for which $2^{\gamma}$ is also transcendental. 
Now let $\alpha = 2^{\gamma}$ and $\beta = 1/\gamma$.
