Small quotients of smooth numbers Assume that $N=2^k$, and let $\{n_1, \dots, n_N\}$ denote the set of square-free positive integers which are generated by the first $k$ primes, sorted in increasing order. Question: what is a good lower bound for
$$
\min_{1 \leq \ell_1  < \ell_2 \leq N} ~ \frac{n_{\ell_2}}{n_{\ell_1}}.
$$
Remark 1: Since by assumption $n_{\ell_2} > n_{\ell_1}$ for $\ell_2 > \ell_1$, a trivial lower bound is 1.
Remark 2: By the prime number theorem, the largest number $n_N$ is of size roughly $e^{c k \log k}$ for some $c$. Thus an obvious lower bound is something like
$$
\frac{e^{c k \log k}}{e^{c k \log k} -1} \approx 1 + \frac{1}{e^{c k \log k}}.
$$
The questions is if an essential improvement of this simple lower bound is possible. In particular, it would be nice to get a lower bound of size roughly
$$
1 + \frac{1}{e^{c k}}.
$$
(All constants $c$ in these statements are generic positive numbers, their specific values are not of interest for me. Also, I don't care about small values of $k$ or $N$, but about an asymptotic result.)
Edit: It would already be very helpful to know that 1% (or any other fixed percentage) of all possible quotients of consecutive numbers $n_{\ell+1}/n_\ell$ satisfy the desired lower bound.
 A: (New part at the bottom):
Here is a long comment:
A more general question was answered by Tijdeman:
1) On integers with many small prime factors (Compositio, 26 (3), 1973,
319-330.
http://archive.numdam.org/article/CM_1973__26_3_319_0.pdf
But sees also 
2) Tijdeman On the maximal distance between integers composed of small primes,
Compositio 28 (2), 1974, 159-162.
http://archive.numdam.org/article/CM_1974__28_2_159_0.pdf
He studies gaps between S-units, i.e. integers multplicatively generated
by (any) $k$ prime factors. Your sequence of square-free integers
generated by the first $k$ primes is a subset of Tijdeman's sequences, hence
a lower bound for Tijdeman sequences gives a lower bound for your problem.
Using linear forms in logarithms, (Baker theory), Tjdeman proves (Corollary on
page 321): There exists an effectively computable constant $C=C(k)$
such that
$$n_{i+1}-n_i > \frac{n_i}{(\log n_i)^{C(k)}}, \text{ for } n_i \geq 3.$$
Dividing by $n_i$ gives that
$\frac{n_{i+1}}{n_i} > 1+ \frac{1}{(\log n_i)^{C(k)}}\geq 1+ \frac{1}{(2k\log k)^{C(k)}}$,
(where the $2$ can certainly be improved with some care on the estimate on the
largest $n_i$).
Therefore 
$$
\min_{1 \leq \ell_1  < \ell_2 \leq N} ~ \frac{n_{\ell_2}}{n_{\ell_1}} 
> 1+ \frac{1}{(2k\log k)^{C(k)}}.$$
If this bound is of any use for you seems to depend on $C(k)$. It seems unlikely that the effectively computable $C(k)$ has been computed, with any useful value.
EDIT:
In a series of papers R.R. Hall, P. Erdös, H. Maier and G. Tenenbaum studied questions of the "propinquity" of divisors.
Maybe the following is useful:
MR0554398, Erdős, P.; Hall, R. R.
The propinquity of divisors.
Bull. London Math. Soc. 11 (1979), no. 3, 304–307. 
Theorem 1 (or short excerpt of MR entry): Let $\epsilon>0$ be fixed, and let
$$\eta(x)=\exp(−(1+\epsilon)(\log 3)(2\log \log x)^{1/2}(\log \log \log \log x)^{1/2}),$$
$$\theta (x,d)=\eta(x)(\log d)^{1−\log 3}$$. Then the number of integers $n$ not exceeding $x$, and having divisors $d,d′$ with $$d<d′<d(1+\theta(x,d))$$ is $o(x)$.
The sequence of the product of the first $k$ prime factors is of course a quite thin sequence, but it should not behave too irregularly, and maybe ideas of the proof can be used.
If one makes the above function above slightly larger, then
MR0739628 Maier, H.; Tenenbaum, G. On the set of divisors of an integer.
Invent. Math. 76 (1984), no. 1, 121–128 shows that
almost all $n$ do have such (multiplicatively) close divisors.
A: Let $P_k$ be the product of the $k$ first prime numbers, and for any integer $n$, we denote by $(d_i)_{1 \leq i \leq \tau(n)}$ the increasing sequence of its divisors.
The set $\{n_1, ... n_N \}$ of square-free positive integers which are generated by the first $k$ primes that you are considering is nothing else than the set of the divisors of $P_k$. Thus we can use this notation:
$$\delta_k = \min_{\ell_1<\ell_2} \left(\frac{n_{\ell_2}}{n_{\ell_1}} - 1\right)
 = \min_{\ell} \left(\frac{n_{\ell+1}}{n_\ell} - 1\right) = \min_{d_i | P_k,\  1 \leq i < 2^k} \left(\frac{d_{i+1}}{d_i} - 1\right).$$
We have, for $ \epsilon > 0$,
$$(2^k-1) \Bigg( \min_{d_i | P_k,\  1 \leq i < 2^k} \left(\frac{d_{i+1}}{d_i} - 1\right)\Bigg)^{1+\epsilon} \leq \sum_{1 \leq i < 2^k} \left(\frac{d_{i+1}}{d_i} - 1\right)^{1+\epsilon} \leq C,$$
with an absolute constant $C$, uniformly for all $k$, by using Théorème 1 of Sur un problème extrémal en arithmétique, G. Tenenbaum, Ann. Inst. Fourier, Grenoble (1987).
i.e. it is proved that, for every $\epsilon > 0$,
$$\delta_k \ll 2^{-k(1-\epsilon)}$$
Note 1 : It doesn't give the lower bound you were looking for, but it confirms that $\delta_k$ has an upper bound of the order you expected, i.e. $e^{-c\ k}$.
Note 2 : I have applied Théorème 1 with the function $h(u) = u^{1+\epsilon}$, but the article describes (see page 3) a wider class of function that could be used.
Note 3 : For a lower bound, the best I can prove is (do you want me to post a proof for this?):
$$\delta_k \gg e^{- \frac{1}{2} k \log k}$$
