If we take the primes $p_k < n$, and raise them to the highest power possible such that $(p_k)^{r_k} \le n$, what is the lower bounds on $\prod{ (p_k)^{r_k} }$? In other words, what are the asymptotics of this function?
For example, if $n=10$, $2^3 \le 10$, $3^2 \le 10$, $5^1 \le 10$, and $7^1 \le 10$. This product is $2^3 3^2 5^1 7^1 = 2520$. So the function at $10$ is $2520$. Again, how closely can we bound this function?
Note that the Chebyshev $\psi\left(x\right) $ function is defined as $$\psi\left(n\right)=\sum_{p^{k}\leq n}\log\left(p\right) $$ where the sum is for all powers of prime less or equal than $x $. It is well known that $$\psi\left(n\right)=e^{\textrm{lcm}\left(1,2,\dots,n\right)}. $$ Let $m\left(p\right)$ the maximum power of $p $ such that $p^{m\left(p\right)}\leq n$. Then is sufficient to note that $$m\left(p\right)=\left\lfloor \log_{p}\left(n\right)\right\rfloor $$ hence our product is $$\log\left(\prod_{p\leq n}p^{m\left(p\right)}\right)=\sum_{p\leq n}m\left(p\right)\log\left(p\right) $$ $$=\sum_{p\leq n}\left\lfloor \log_{p}\left(n\right)\right\rfloor \log\left(p\right)=\psi\left(n\right) $$ so $$\prod_{p\leq n}p^{m\left(p\right)}=e^{\psi\left(n\right)}=\textrm{lcm}\left(1,2,\dots,n\right).$$ Using the Prime Number Theorem we know that $$\psi\left(n\right)\sim n $$ so we have the asymptotic $$\prod_{p\leq n}p^{m\left(p\right)}\sim e^{n}.$$