least common multiple of $\{1,2,...,n\}$ is bigger than $2^{n-1}$

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.

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 $$ \operatorname{lcm}(1,2,\dots,n) \geq n \binom{n-1}{\lfloor (n-1)/2 \rfloor} \geq \sum_{k=0}^{n-1} \binom{n-1}{k}=2^{n-1} $$

(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: $$ \begin{array}{|c|c|c|} \hline n & \psi(n) & (n-1)\ln 2 \\ \hline 2 & 1.79176 & 0.693147 \\ 3 & 2.48491 & 1.38629 \\ 4 & 4.09434 & 2.07944 \\ 5 & 4.09434 & 2.77259 \\ 6 & 6.04025 & 3.46574 \\ 7 & 6.73340 & 4.15888 \\ \end{array} $$