The inequality holds for $n = 2,3,4,5,7,9$ and does not hold for $n = 1,6,8,10,11,12$. Assume in the rest that $n \ge 13$. By definition of $\kappa(n)$, there exists exactly $n-1-\kappa(n)$ prime numbers not greater than $n$; these are the primes dividing $\text{lcm}(1,2,\dots,n)$. We have $n-1-\kappa(n) \ge 6$ as $n \ge 13$. Let $q_1, \dots, q_{n-1-\kappa(n)}$ be all prime powers in the canonical representation of $\text{lcm}(1,2,\dots,n)$. W.l.o.g., $q_1 > q_2 > \dots > q_{n-1-\kappa(n)}$. As at least 5 numbers among $q_1, \dots, q_6$ are odd, in the case of odd $n$ we have $q_1q_2q_3q_4q_5q_6 \le n(n-1)(n-2)(n-4)(n-6)(n-8)$ and in the case of even $n$ similarly $q_1q_2q_3q_4q_5q_6 \le n(n-1)(n-3)(n-5)(n-7)(n-9)$. But since $(n-3)(n-5)(n-7)(n-9) < (n-2)(n-4)(n-6)(n-8)$ and $n(n-8) < (n-3)(n-5)$, we anyway obtain
$$q_1q_2q_3q_4q_5q_6 < (n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = \frac{(n-1)!}{(n-7)!}$$
As $q_6 < n-6$, the inequality $q_i < n-i$ holds for every $i > 6$, whence
$$q_7 \cdots q_{n-1-\kappa(n)} \le (n-7) \cdots (\kappa(n)+1) = \frac{(n-7)!}{(\kappa(n))!}$$
Consequently, $\text{lcm}(1,2,\dots,n) = q_1q_2\cdots q_{n-1-\kappa(n)} < \frac{(n-1)!}{(\kappa(n))!}$, contradicting the original inequality. Hence the inequality holds for $n = 2,3,4,5,7,9$ only.