Is there a simple proof for the irrationality or transcendence of $e^{q}$ where $q \in \mathbb{Q}$? This link gives the cases of $n \in \mathbb{N}$*
Proving the irrationality of $e^n$
What about cases of rational exponents $q \in \mathbb{Q}$?
 A: Write $q=a/b$. 
If $e^q$ is rational, then $e^a=(a^q)^b$ is rational.
Similarly, if $e^q$ is algebraic, then $e^a=(e^q)^b$ is algebraic.
So if you know that $e$ is transcendental, you get it for all $e^q$.
A: Irrationality of $e^q$ for nonzero $q\in\mathbb{Q}$ follows immediately from irrationality of $e^n$ for positive integers $n$.  Indeed, suppose $q=\frac{a}{b}$ where $a$ and $b$ are integers.  We then have $(e^q)^b=e^a$, so if $e^q$ were rational, $e^a$ would also be rational.  Taking the reciprocal if $a<0$, we get that $e^{|a|}$ is rational.  But if $q\neq 0$, then $|a|$ is a positive integer, so $e^{|a|}$ is irrational.  Thus $e^q$ must be irrational as well.
The exact same argument works for transcendence, replacing "rational" with "algebraic".
A: Let suppose $e$ is a rational number, and it can be written by $2$ integer which are pride to other ;
$$e=p/q \quad p,q\in\mathbb Z\quad (p\nmid q)$$
$$e=\lim\limits_{n\to\infty}\displaystyle\sum_{i=0}^n\dfrac{1}{i!}$$
To gain a national number, let's multiply each side by $q!$
$$eq!=\lim\limits_{n\to\infty}\displaystyle\sum_{i=0}^n\dfrac{q!}{i!}=\underbrace{\sum_{i=0}^q\dfrac{q!}{i!}}_R+\underbrace{\lim\limits_{n\to\infty}\displaystyle\sum_{i=q+1}^n\dfrac{q!}{i!}}_Q$$
Since $eq!$ is a national number, right side of the equation is a national number, as well. Moreover, $R$ is a national number too, so $Q$ must be a national number, but.. 
$$Q=\lim\limits_{n\to\infty}\displaystyle\sum_{i=1}^n\dfrac{q!}{(i+q)!}=\lim\limits_{n\to\infty}\displaystyle\sum_{i=1}^n\dfrac{1}{(1+q)(2+q)...(i+q)}$$
Due to this inequality,
$$i,q \in \mathbb N\\ (1+q)(2+q)...(i+q)>(1+q)^i$$
and we now see that this assumption is impossible, because;
$$0<Q<\lim\limits_{n\to\infty}\sum_{i=i}^n\dfrac1{(1+q)^i}=\dfrac{1}{1-\frac{1}{1+q}}-1<1$$
There isn't any element of national numbers that can be $$(n\in\mathbb N>0)\quad\wedge\quad(n\in\mathbb N<1)$$ $\Box$
A: I assume you have seen it proven that $e $ is transcendental.
So $m^r*e^s-n^r \ne 0$ for any integers $r,s,n,m;s\ne 0$.
So $e^{s/r}\ne n/m $ for any rational $\frac sr,\frac nm $.
