Must every length-$p_1\cdots p_k$ interval contain an integer with $>k$ prime factors? (Erdős #891)

Let $2=p_1<p_2<\cdots$ be the sequence of primes and fix an integer $k\geq 2$. Is it true that, for all sufficiently large $n$, the interval $$[\,n,\ n+p_1p_2\cdots p_k\,)$$ of $p_1\cdots p_k$ consecutive integers must contain an integer with more than $k$ prime factors (counted with multiplicity)? The first open case is $k=2$, where $p_1p_2=6$: is it true that every interval of $6$ sufficiently large consecutive integers contains one with at least $3$ prime factors?

Acceptance. FULLY RESOLVES (proof-shaped): a complete written or machine-checkable proof that for each fixed $k\geq 2$ every interval $[n,n+p_1\cdots p_k)$ with $n$ large contains an integer with more than $k$ prime factors, or a disproof producing, for some $k$, infinitely many intervals of length $p_1\cdots p_k$ in which every integer has at most $k$ prime factors. Settling the single case $k=2$ (every $6$ large consecutive integers include one with $\geq 3$ prime factors) counts as fully resolving that case. ADVANCES (checkable): prove the statement with the interval length improved below Schinzel's $p_1\cdots p_{k-1}p_{k+1}$ toward the target $p_1\cdots p_k$, with proof; establish a conditional resolution under a clearly stated hypothesis (e.g. a quantified prime $k$-tuples / Dickson-type input), flagged as conditional; or a certified computation reporting, over a documented range and for small $k$, the longest run of consecutive integers each with at most $k$ prime factors, bounding how large $n$ can be while a length-$p_1\cdots p_k$ counterexample persists. Deliver the proof/formalisation or the reproducible search code with certified output.