I was surprised that the numbers $\pi$, $\ln{(2)}$, $\zeta{(2)}$, and $\zeta{(3)}$ can be shown to be irrational in what seems to be "three-lined proofs" (as identified here on Overflow: [Establishing zeta(3) as a definite integral and its computation.](https://mathoverflow.net/questions/30659/establishing-zeta3-as-a-definite-integral-and-its-computation/30784#30784)).


The idea goes that we suppose, for $j \in \mathbb{N}$, we have a family of non-zero integrals such that $$\int\_0^1 x^j f(x)\ dx = P\_j \xi+ Q\_j $$ where $P\_j, Q\_j \in \mathbb{Q}$, $f(x)$ is arbitrary, and $\xi$ is some number. If $\xi$ is rational, then $\int\_0^1 x^j f(x)\ dx = A\_j/B\_j $ yields a rational expression. Multipying $a\_{nj} \in \mathbb{Z}$ to both sides, and taking the sum from $j = 0$ to $n$, we have 
$$\sum\_{j=0}^n a\_{nj} \int\_0^1 x^j f(x)\ dx = \sum\_{j=0}^n a\_{nj} \frac{A\_j}{B\_j} = \frac{E\_n}{F\_n} $$ where $E\_n, F\_n \in \mathbb{Z}$.


Beuker treated the $a\_{nj}$ coefficients as the coefficients of the *shifted Legendre polynomials* in his proof for the irrationality of $\zeta{(3)}$, where


$$ P\_n{(x)} = \frac{1}{n!} \frac{d^n}{dx^n} (x^n (1-x)^n) = \sum\_{j=0}^n a\_{nj} \, x^j$$


This leads to


$$\int\_0^1 \bigg( \sum\_{j=0}^n a\_{nj} \, x^j \bigg) f(x)\ dx = \int\_0^1 P\_n{(x)}\,f(x) \ dx = \frac{E\_n}{F\_n}$$


If we can show, as $n$ gets larger,


$$ \left| F\_n \int\_0^1 \frac{1}{n!} (x^n (1-x)^n)\, \frac{d^n}{dx^n}f(x) \ dx \right| = |E\_n|\to 0 $$ 


we get a contradiction, since at "some point", $|E\_n|$ is an integer between $0$ and $1$. Thus, $\xi$ must be irrational. Of course, the hard part is finding a suitable function $f(x)$.


Since $f(x)$ has been found for the proofs of the above numbers, has it been shown the number $e$ can be proved irrational using the above method (implying has such an $f(x)$ been found)? 



Question on the irrationality of $e$
One has $$\int\_0^1 x^k e^x dx = A\_ke+B\_k \to 0$$ as $k\to \infty$, $A\_k,B\_k\in \mathbb{Z}$, by integration-by-parts and induction. If $e=a/b$, then $(A\_k a+B\_kb)/b \to 0$, so $A\_ke+B\_k=0$ for $k$ large, a contradiction. 


