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 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.

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 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.

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 $(8,9,10)$, 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 an "easy" argument.

The question is essentially the same as this one 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.

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 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.