Does there exist any positive integer $n$ such that $e^n$ is an integer (to show $\log 2$ is irrational)?

Question

Does there exist any positive integer $n$ such that $e^n$ is an integer ?

I was in particular trying to prove $\log 2$ is irrational; now if it is rational, then there are relatively prime integers $p,q$ both positive such that $\log2 =p/q$ that is $e^p=2^q$ is an integer. I wanted to reach a contradiction.

Answer 1

No $e$ is transcendantal and if there were integers $n$ and $m$ such that $e^n=m$ and therefore would be algebraic over $\mathbb{Q}$ as a root of the polynomial $X^n-m$. Contradiction.

Answer 2

Assuming that for some couple $(n,m)$ of natural numbers we have $e^n=m$, then $\tanh(n)\in\mathbb{Q}$.

Since $\tanh z=\frac{d}{dz}\log\cosh z$, a continued fraction representation for $\tanh n$ is given by:

$$ \tanh n = \cfrac{1}{\frac{1}{n}+\cfrac{1}{\frac{3}{n}+\cfrac{1}{\frac{5}{n}+\cfrac{1}{\frac{7}{n}+\cfrac{1}{\frac{9}{n}+\ldots}}}}} \tag{1} $$ and it is not difficult to prove that such expression cannot be equal to some finite standard continued fraction. This leads to $\tanh n\not\in\mathbb{Q}$, hence to $e^n\not\in\mathbb{N}$ as wanted.