TITLE: Proving the irrationality of $e^n$
QUESTION [17 upvotes]: Let $n$ be a positive integer. I know the traditional proof that $e$ is irrational. How do we show that $e^n$ is irrational in some sort of similar line? I am of course assuming it is but I would be astounded if not. It occurs to me that since $e$ is transcendental, of course $e^n$ is irrational, but I don't want to use that fact.
Googling gives me something for $e^2$, but I could not easily find anything for $e^3$.

REPLY [3 votes]: You can find a pretty direct proof of this result here, in proposition 7.
Here is the essence of the argument: for $r(x)$ a polynomial of degree $k$ we can express $$\int_0^1r(x)e^{z-xz}dx=\frac{F(z,0)e^z-F(z,1)}{z^{k+1}}$$
where $F(z,x)$ is a polynomial in $z$ whose coefficients are successive derivatives of $r$ at $x$ (this is proposition 5). In particular, if $e^p=\frac{a}{b}$ is rational and we take $r(x)=x^n(1-x)^n$ for some large $n$ to be specified later, we can use this formula to estimate $aF(p,0)-bF(p,1)\leq ap^{2n+1}$. However, $aF(p,0)-bF(p,1)$ is clearly positive (by looking at the integral), and from the formula for $F(z,x)$ and the fact that many derivatives of $r(x)$ vanish at $0,1$ we can deduce $aF(p,0)-bF(p,1)$ is integer divisible by $n!$. Hence $n!\leq aF(p,0)-bF(p,1)\leq ap^{2n+1}$, so $n!\leq ap^{2n+1}$ which is contradictory for large $n$.