Exponential extension of $\mathbb{Q}$ A non-trivial exponential function $E:\mathbb{K} \rightarrow \mathbb{K}$ in a field $\mathbb{K}$  is a function such that
\begin{split}
E(x+y)=E(x)E(y) \quad \forall x,y \in \mathbb{K} \\
E(x)=1 \iff x=0
\end{split}
For the exponential function such that $E(1)=a \in \mathbb{K}$ write $E_a(X)=a^x$.
We know that if $\mathbb{K}=\mathbb{Q}$ such a function can not be defined as $a^{\frac{m}{n}} $ 
can be irrational, and  $E_a(x) \notin \mathbb{Q}$.
Call exponential extension of $\mathbb{Q}$ an extension of $\mathbb{Q}$ in which we can define an exponential function. We know that  $\mathbb{R}/\mathbb{Q}$ and $\mathbb{C}/\mathbb{Q}$ are such exponential extension. But it can be shown that there is no exponential extension $\mathbb{E}/\mathbb{Q}$ whith $ \mathbb{E}\subset \mathbb{R}$ and $\mathbb{E}\ne \mathbb{R}$?
I can not find such a demonstration.
(Sorry for my bad English).
 A: Let be
$$\Bbb E_0=\Bbb Q$$
$$B_1=\{2^r\,|\, r\in\Bbb E_0\}$$
$$\Bbb E_1=\Bbb E_0(B_1)$$
$$B_2=\{2^r\,|\, r\in\Bbb E_1\}$$
$$\Bbb E_2=\Bbb E_1(B_2)$$
$$\cdots$$
$$\Bbb E=\bigcup_{n\in\Bbb N}\Bbb E_n$$
and $\Bbb E\ne\Bbb R$ because is countable.
A: Thank you very much! 
If I have well understood: for each $a \in \mathbb{Q}$ we can build a field 
$$ 
\mathbb{E}_a=\bigcup_{n=0}^{\infty}\mathbb{E}_{a,n} 
$$
such that $E: \mathbb{E}_a \rightarrow  \mathbb{E}_a$, $E_a(x)=a^x \quad \forall x \in  \mathbb{E}_a$ is a well defined exponential function.
Now the problem becomes: 
the field closures of the sets 
$$
\mathbb{E}=\bigcup_{a \in\mathbb{Q}}\mathbb{E}_a \varsubsetneq \mathbb{C} \qquad \mbox{ and } \qquad
\mathbb{E}^+=\bigcup_{a \in\mathbb{Q}^+}\mathbb{E}_a \varsubsetneq \mathbb{R}
$$
are are proper subsets of $\mathbb{C}, \mathbb{R}$ ?
And I get a further question: the Napier's constant  $e$ is an element of $\mathbb{E}^+$?
And, more generally, there is a method to determine if a transcendental number $ \in \mathbb{E}$?
