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 $n$.