This is a really awesome question. I haven't spent much time thinking about this problem, so I'm sure you have better intuition on it than me. I have a heuristic argument that appears to suggest that what you want is indeed true for large $n$. My reasoning is as follows.
Let $P(i)$ denote the largest prime factor of a positive integer $i$, and let $S(x)$ be defined by
$$S(x) = \sum_{2 \leq i \leq x} P(i).$$
It is a result of Krishnaswami Alladi and Paul Erdos (see their paper: Pacific J. Math. 71(1977) 275-294) that
$$S(x) \sim \frac{\pi^2}{12}\frac{x^2}{\log x},$$
where by ``$\sim$'' I mean asymptotic equivalence (approximate order of growth -- remember, this is a heuristic argument, not meant to be precise). Then the sum of the largest prime factors of the numbers $n+2, \dots, n+k$ is given approximately by $$\sum_{i = n+2}^{n+k} P(i) = \sum_{i = 2}^{n+k} P(i) - \sum_{i = 2}^{n+2} P(i) \sim \frac{\pi^2}{12}\left(\frac{(n+k)^2}{\log(n+k)} - \frac{(n+2)^2}{\log(n+2)}\right).$$
To leading order in $n$, the last expression above is approximately
$$\frac{\pi^2k}{6}\frac{n}{\log n}.$$
Now, if your claim is true, that the product $(n+2)\cdots(n+k)$ has a prime factor greater than $k$ for all $n \geq k \geq 3$, then the above approximate sum should be bigger than $k$ times the number of factors being multiplied, which is $n+k - (n+2) + 1 = k - 1$, so again working to leading order, you end up getting the approximate inequality
$$\frac{\pi^2k}{6}\frac{n}{\log n} \gg k(k-1),$$
and this inequality actually does hold for large $n$. This intuition tells me that your conjecture is true!
To conclude, let me mention where the above argument is not rigorous. I used the result of Alladi and Erdos in a subtraction, but perhaps the error term on their result is so large that such an argument is invalid. I also threw out a lot of non-leading terms in places that may have actually been important.
An alternative strategy would be to try improving upon Sylvester's Theorem; look up a proof and see if you can optimize the methods!
