$\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$ I can prove that $\text{lcm}(1,2,3,\ldots,n)\geq 2^{n-1}$.
Newly, i read in a paper that  for $n\geq 7$ we have:
$$\text{lcm}(1,2,3,\ldots,n)\geq 2^n$$
Can you prove it?
(This inequality is an interesting inequality. For example with this inequality can find a lower bound for the number of primes less than $n$.)
 A: The proof for this, is almost the same for proving the bound of $2^{n-1}$. Consider for $1\le m\le n$
$$I_{m,n}=\int_{0}^{1}x^{m-1}(1-x)^{n-m}=\sum_{r=0}^{n-m}\frac{a_r}{m+r}\quad \text{ for some }a_i\in \mathbb{Z}$$
Let $\displaystyle\ell_n=\text{lcm}(1,2,\dots,n)$. See $\ell_nI_{m,n}\in \mathbb{Z}$ for $1\le m\le n$. Now it can be easily seen that $\displaystyle I_{m,n}=\frac{1}{m\dbinom{n}{m}}$. Which will follow from integration by parts or reduction formulae. Then we have:
$$m\dbinom{n}{m}\mid \ell_n\quad \forall \;\;1\le m\le n$$
In general $$n\dbinom{2n}{n}\mid \ell_{2n}\quad \text{ and }\quad (2n+1)\dbinom{2n}{n}=(n+1)\dbinom{2n+1}{n+1}\mid \ell_{2n+1}$$
Now since $\ell_{2n}\mid \ell_{2n+1}$ we have 
\begin{align*}
& n(2n+1)\dbinom{2n}{n}\mid \ell_{2n+1}\\
\implies & \ell_{2n+1}\ge n(2n+1)\dbinom{2n}{n}\ge n\cdot 4^n\ge 2^{2n+2}\quad \text{ for } \quad n\ge 4\end{align*}
The second last inequality holds since $\dbinom{2n}{n}$ is the largest co-efficient in the expansion of $(1+x)^{2n}$. Hence $(1+1)^{2n}\le \dbinom{2n}{n}(2n+1)$.
Also $$\ell_{2n+2}\ge \ell_{2n+1}\ge 2^{2n+2}\quad \text{ for }\quad n\ge 4$$
Hence we have proved $$\ell_n\ge 2^{n}\quad \text{ for }\quad n\ge 9$$
Smaller cases are checked by hand. Thus we are done.
Also about the fact that $\log \ell_n\approx n$. This is a sketch.
Take some prime $p$ dividing $\ell_n$. Now $\nu_p$ be the highest exponent of $p$ in $\ell_n$. Then $p^{\nu_p}\mid k$ for some $1\le k\le n$, then $p^{\nu_p}\le n\implies \nu_p\le \frac{\log n}{\log p}$
Now 
\begin{align*}
&\ell_n=\prod_{p\le n}p^{\nu_p}\le \prod_{p\le n}p^{\frac{\log n}{\log p}}\\
\implies & \log \ell_n\le \sum_{p\le n}\left(\frac{\log n}{\log p}\cdot \log p\right)=\pi(n)\log n\approx n\quad \text{for large }n\end{align*}
where the last statement holds due to the prime number theorem.
