Let $n$ be a positive integer, and $f(2n+1)$ be equal to $lcm[1,2,\ldots,2n+1]$. I am supposed to show that $\log(f(2n+1))\ge 2n\log 2$ by using the fact that $e^{f(2n+1)}\int\limits_{0}^{1} x^n(1-x)^n~dx\in\mathbb N$.
Suppose that I accept the fact in faith. One notices that $x(1-x)=1/4-(1/2-x)^2\le 1/4$ whenever $x\in[0,1]$. Thus $$I:=\int\limits_{0}^{1} x^n(1-x)^n~dx\le \int\limits_{0}^{1}(1/4)^n~dx=2^{-2n},$$ and so $1\le e^{f(2n+1)}I\le e^{f(2n+1)}2^{-2n}$, i.e., $e^{f(2n+1)}\ge 2^{2n}$. The rest follows by taking natural logarithm on both sides. Thus if I can show the fact I will be done. Here is where I am stuck. By induction, it can be calculated rather easily that $$I=\frac{n\cdot (n-1)\cdots 1}{(2n)(2n-1)\cdots (n+1)}\cdot \frac{1}{2n+1},$$ which according to the fact must be in $e^{-f(2n+1)}\mathbb N.$ Any help is appreciated. Thank you!