Take $k$ consecutive composite integers from a prime gap. What is known about the largest prime divisor of their product?


It seems to me that except for the triplet $(8,9,10)$ and the pair $(8,9)$ , this largest prime divisor is always larger than $2k$, but I could not find an elementary (my level) proof. 


As $k$ grows it seems that the sharpness of $2k$ as lower bound is lost, therefore I expect that there should exist a more suitable lower bound, from which an "easy" argument for the minoration by $2k$ would follow. 


The question is essentially the same as [this one](https://math.stackexchange.com/questions/1495443/2k-as-a-lower-bound-for-the-largest-prime-divisor-of-binomial-coefficient-bi) from MSE but it did not get any answer. I could only find messy $k$-specific arguments for $k$ up to $4$ which I hope are correct. 



What is known about the largest prime divisor of the product of $k$ consecutive integers?
There have been several investigations into the largest prime factor of a product of consecutive integers; the Sylvester--Schur theorem is an early example. Here is a survey by Shorey and Tijdeman: <https://www.math.leidenuniv.nl/~tijdeman/shoretij.pdf>


In 


Laishram, Shanta(6-TIFR-SM); Shorey, T. N.(6-TIFR-SM)
The greatest prime divisor of a product of consecutive integers. 
Acta Arith. 120 (2005), no. 3, 299–306


it is shown that $$ P(n(n+1) \cdots (n+k-1)) > 2k $$ as long as $n > \max\{k+13, \frac{279}{262}k\}$. 


Note that for large $k$, the lower bound condition on $n$ is implied by your hypothesis that all of $n, n+1, \dots, n+k-1$ are composite; so this answers your question for large $k$. (And for bounded $k$, it reduces checking your conjecture to a finite computation.) 


There is more good news: the MathSciNet review records that "the proof is elementary". 


