Proving $e$ is irrational using a Beukers-like integral We know the following, for some integers $a_n,b_n$ and $n$ where $n\geq0$:
\begin{align}
&I_n = \int_0^1 x^n(1-2x)^n e^x dx = a_ne+b_n \\
&|I_n|= \left\lvert \int_0^1 x^n(1-2x)^n e^x dx  \right\rvert \leq \left( \frac{1}{8} \right)^n(e-1)
\end{align}
therefore, as $n\to\infty$ the integral $I_n$ tends to zero. So if we assume $e=p/q$ then we would have a integer that tends to zero which is not possible, unless the integer is $0$ itself. So, if we prove that $I_n$ is not zero, then this concludes the proof that $e$ is irrational.
But how to prove that $I_n \neq0$ for all integer $n\geq0\:?$
PS: I know that if we changed the polynomial inside $I_n$ it would be easier to show that $I_n \neq 0$ for all $n$, but I'm interested in this case in particular.
EDIT: I believe the estimation for $I_n$ is wrong. I estimated it by the following way:
\begin{align}
&x(1-2x)\leq\text{max}[x(1-2x)] = 1/8\\
&x^n(1-2x)^n\leq  \left( 1/8 \right)^n\\
& x^n(1-2x)^n e^x \leq  \left( 1/8 \right)^n e^x\\
& \int_0^1 x^n(1-2x)^n e^x dx\leq \left( 1/8 \right)^n \int_0^1 e^xdx\\
& \int_0^1 x^n(1-2x)^n e^x dx\leq \left( 1/8 \right)^n (e-1)
\end{align}

The case is trivial when $n$ is even ($I_n$ always $>0$), so we just assume $n$ is odd. Write
$$I_n=\underbrace{\int_0^{1/2} x^n(1-2x)^n e^x~\mathrm  dx}_{J_1}+\underbrace{\int_{1/2}^1 t^n(1-2t)^n e^t~\mathrm  dt}_{J_2}.$$
By considering the area under the curve, $J_1>0$ while $J_2<0$. 

Claim: $-J_2> J_1$ when $n$ is odd ($\Leftrightarrow I_n=J_1+J_2<0$).

Proof: Substitute $x=t-\frac12$ in $-J_2$,
\begin{align*}
-J_2=-\int_0^{1/2}\left(x+\frac12\right)^n(-2x)^n e^{x+1/2}~\mathrm dx&=\underbrace{(-1)^{n+1}}_{=1}e^{1/2}\int_0^{1/2}\left(2x+1\right)^n x^n e^x~\mathrm dx\\
(\text{Using the fact $e>1$})\quad&>\int_0^{1/2}\left(2x+1\right)^n x^n e^x~\mathrm dx\\
(2x+1>1-2x\text{ when } x>0)\quad&>\underbrace{\int_0^{1/2}\left(1-2x\right)^n x^n e^x~\mathrm dx}_{J_1}\\
\end{align*}
and we are done.
