This question was asked at [MSE](https://math.stackexchange.com/questions/1465274/is-2n-1-finitely-many-times-the-product-of-consecutive-primes) but recieved no attention at all. 
Here it is: 


Are there finitely many $(n,k) \in \mathbb{N}^2$ with $2^n-1=p\_1p\_2\cdots p\_k$ ? 


$p\_1=3,p\_2=5 , ...,p\_k$ are consecutive odd primes in ascending order.  

An example is when $n=4, k=2$:  

$2^4-1=3\cdot 5=p\_1p\_2$  

Are there finitely many $n$?  

I tried to use [Zsigmondy's theorem](https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem)
without success.  

Thanks in advance!



Is $2^n -1$ finitely many times the product of consecutive primes?
$\let\dvds\mid$Yes. By Zsigmondy's theorem, $2^{12}-1$ has some prime divisor $p\_s$ not dividing $2^i-1$ for $i<12$ (in fact, $p\_s=13$). Now, if $2^n-1=p\_1p\_2\dots p\_k$ with $k\geq s$, then $p\_s\dvds 2^n-1$, so $12\dvds n$ and hence $3^2\dvds 2^n-1$, which is impossible. Thus only the cases with $k<s$ are left.


