least common multiple of $\{1,2,...,n\}$ is bigger than $2^{n-1}$ 
The least common multiple of $\{1,2,...,n\}$ is greater than $2^{n-1}$ for any $n \ge 3$.

I found this in a MATHEMATICA book, but I don't know how to prove this. Can you help me?
[Edit: This thread has a discussion of an asymptotic stronger result, but that relies on the Prime Number Theorem. What else is known about this? JL]
 A: This paper proves the identity
$$
\operatorname{lcm}(1,2,\dots,n)=n \operatorname{lcm}\left(\binom{n-1}{0}, \binom{n-1}{1},\dots,\binom{n-1}{n-1}\right)
$$
by computing the number of factors of $p$ which appear in each expression, for all primes $p$.
From this it follows that
$$
\operatorname{lcm}(1,2,\dots,n) \geq n \binom{n-1}{\lfloor (n-1)/2 \rfloor} \geq \sum_{k=0}^{n-1} \binom{n-1}{k}=2^{n-1}
$$
A: Here is a neat proof:
Consider the product
$$\text{lcm}(1, 2, \dots, 2n + 1)\int_0^1 (x)^n(1 - x)^n \text{d}x$$
This is an integer, because expanding using the binomial theorem and integrating each part gives an integer.  Additionally, the integral is less than
$$\int_0^1 \left(\frac14\right)^n \text{d}x = \frac{1}{4^n}$$
so
$$\text{lcm}(1, 2, \dots, 2n + 1) > 4^n = 2^{2n}$$
which proves the desired inequality for odd values.
To prove it for even values, consider the product
$$\text{lcm}(1, 2, \dots, 2n)\int_0^1 (x)^{n}(1-x)^{n-1} \text{d}x$$
This product is also an integer.  Additionally, the integral is less than
$$\int_0^1 x\left(\frac14\right)^{n-1}\text{d}x = \frac{1}{2\cdot4^{n-1}}$$
so we arrive at
$$\text{lcm}(1, 2, \dots, 2n) > 2\cdot4^{n-1} = 2^{2n - 1}$$
which proves the desired result for even values.
A: (Updated due to lhf's comment).
In a few words,
$$
LCM(1,2,...,n) = e^{\psi(n)},
$$
where $\psi(n)$ is second Chebyshev function (see formula here).
In fact, it is enough to prove that
$$
\psi(n) > (n-1)\ln 2 \approx 0.693(n-1).\tag{1}
$$
Function $\psi(n)$ has asymptotic $\psi(n) \sim n$.
Using lower bound for $\psi(n)$
$$
\psi(n)>0.916n−2.318\tag{2}
$$
(see discussion here and paper here, Lemma $2$, p.$179$)
we get $(1)$ immediately for $n\ge 8$.
And it remains to check $(1)$ manually for $n=1,2,...,7$.
It is shown in this table:
$$
\begin{array}{|c|c|c|}
\hline
n & \psi(n) & (n-1)\ln 2 \\ \hline
2 & 1.79176 & 0.693147 \\
3 & 2.48491 & 1.38629 \\
4 & 4.09434 & 2.07944 \\
5 & 4.09434 & 2.77259 \\
6 & 6.04025 & 3.46574 \\
7 & 6.73340 & 4.15888 \\
\end{array}
$$
