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Let $S(k)$ denote the smallest integer such that there exists a k-term arithmetic progression of primes among the integers $[1,S(k)]$. Green and Tao have an unpublished note that gives a very large upper bound for $S(k)$. Conversely, an elementary argument (sketched below) gives the lower bound $S(k) > (k-1)e^{k}$. My question is, can this lower bound be improved at all?

By considering how the AP reduces mod $p$, it is easy to see that if $q$ is the step of a $k$-term AP of primes then every prime less than $k$ must divide $q$. Thus we have that $S(k) > (k-1) \prod_{p_{i} \leq k } p_{i}$. By taking logarithms and applying the prime number theorem we have that $\prod_{p_{i} \leq k } p_{i} ~ e^{k}$. This gives $S(k) > (k-1)e^{k}$.

Update: As Thomas Bloom points out, the inequality I record isn't true for small $k$. I should have written something more along the lines of $S(k) >(k-1)e^{k-o(k)}$ or $S(k) = \Omega((k-1)e^{k})$ since we know that $psi(x) > x$ infinitely often. The spirit of my question is, are there any arguments (presumably using more sophisticated machinery) that produces a better lower bound (here, I'll accept either bounds that hold for all k, or just infinitely many k).

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