Does this conjecture have a name? What about a counterexample?:
$$ \forall n,k \in \mathbb{N}, k \gt 1, \exists d \in (kn-n,kn] \text{ s.t. } d \perp n! $$
An equivalent statement is this:
Take a Sieve of Eratosthenes-like list of any fixed width $n$ and sieve out multiples of the primes $\leq n$. While keeping the width fixed, let the length of the list extend to infinity. In every row there is an unsieved number (a number that is coprime to all the primes $\leq n$).
Note that for $k \leq n+2$, $d\perp n!$ is necessarily prime. Bertrand's Postulate sets $k=2$. Legendre's conjecture is about the union of rows $n+1$ and $n+2$. Opperman's conjecture is each of these rows individually. This generalization would extend Opperman to any row whatsoever (except the first).
Results to date: Erdos et. al. proved Bertrand. Legendre and Opperman are open. Prime sieves like this one are cyclic in nature, with the distribution of primes repeating every $\prod^{n}_{p \perp n} p$ rows. Ive checked by hand all possible rows in all the possible widths $\leq$ 7.
Examples of sieves:
$n = 5$:
\begin{array}{|c|c|c|c|c|} \hline 1 & \color{grey}{2} & \color{grey}{3} & & \color{grey}{5} \\ \hline & 7 & & & \\ \hline 11 & & 13 & & \\ \hline & 17 & & 19 & \\ \hline & & 23 & & \\ \hline & & & 29 & \\ \hline 31 & & & & \\ \hline & 37 & & & \\ \hline 41 & & 43 & & \\ \hline & 47 & & 49 & \\ \hline & & 53 & & \\ \hline & & & 59 & \\ \hline 61 & & & & \color{white}{65} \\ \hline \end{array}
$n = 6$:
\begin{array}{|c|c|c|c|c|c|} \hline 1 & \color{grey}{2} & \color{grey}{3} & & \color{grey}{5} & \\ \hline 7 & & & & 11 &\\ \hline 13 & & & & 17 &\\ \hline 19 & & & & 23 &\\ \hline & & & & 29 &\\ \hline 31 & & & & &\\ \hline 37 & & & & 41 &\\ \hline 43 & & & & 47 &\\ \hline 49 & & & & 53 &\\ \hline & & & & 59 &\\ \hline 61 & \color{white}{62} & \color{white}{63} & \color{white}{64} & \color{white}{65} & \color{white}{66}\\ \hline \end{array}
Your conjecture does not hold for $n=8, k=25$ as all $8$ numbers $$(201, 202, 203, 204, 205, 206, 207, 208)$$ have factors less than $8$, which are $(3, 2, 7, 2, 5, 2, 3, 2)$ respectively.
For prime $n$, smallest counterexample is $n=13, k=168$ because all $13$ numbers$$2185, 2186, \cdots, 2197$$have factors less than or equal to $13$, which are $(5, 2, 3, 2, 11, 2, 7, 2, 3, 2, 5, 2, 13)$.
I thought your conjecture is true before running a program because intuitively $\prod_{p_i\le n} (p_i-1)/p_i$, which is the ratio of numbers left on each stage of Sieve of Eratosthenes decreases much slower than the growth of $n$.