$2^n$-1 consisting only of small factors I've checked the factorization of $2^N - 1$ up through N = 120 for the largest prime factor, and it looks like the largest value of N where $2^N-1$ has a largest prime factor under 2500 is N = 60 (largest prime factor = 1321). As N gets larger, the largest prime factors get larger, even for the "abundant" numbers like 96, 108, and 120.
Is there a way to prove that no value of N > 60 exists such that the largest prime factor of $2^N - 1$ is less than 2500?
 A: It is true that if $N > 60$, then $2^{N} - 1$ has a prime factor $> 2500$.
Here's another approach. First, observe that every prime factor of $2^{p} - 1$ is $\equiv 1 \pmod{p}$. Combining this with the observation that if $a | b$, then $2^{a} - 1 | 2^{b} - 1$, we see that if $2^{N} - 1$ has all prime factors $\leq 2500$, then all prime factors of $N$ are $< 2500$. Checking these primes, we see that $2^{p} - 1$ has a prime factor $> 2500$ unless $p = 2, 3, 5, 7, 11$ or $29$. Hence if all the prime factors of $2^{N} - 1$ are less than $2500$,
then all prime divisors of $N$ are in the set $S = \{ 2, 3, 5, 7, 11, 29 \}$.
We find that $65537$ is a prime factor of $2^{32} - 1$ and this means that 
$N$ cannot be a multiple of $32$ if $2^{N} - 1$ has all prime divisors $< 2500$.
Similar arguments show that $N$ cannot be a multiple of $3^{3}$, $5^{3}$, $7^{3}$, $11^{2}$ or $29^{2}$. This implies that $N$ divides $56271600$, and checking all such divisors, we see that $N = 60$ is the largest possible.
A: A recent breakthrough on this problem is the work of Cam Stewart (the paper has appeared in Acta Mathematica). Proving a conjecture of Erdos, Stewart shows that 
the largest prime factor of $2^n-1$ is at least 
$$ 
n \exp\Big( \frac{\log n}{104 \log \log n}\Big), 
$$ 
if $n$ is large enough.  He also gives many related earlier results: for example a result of Schinzel which shows that the largest prime factor of $2^n-1$ is at least $2n+1$ for $n\ge 13$, and discusses other work conditional on GRH or abc.  
A: Zsigmondy's theorem implies that when $n \ge 7$, $2^n - 1$ has a prime divisor not dividing $2^k - 1$ for any $k < n$. So pick $n_0$ large enough that every prime number less than $2500$ divides $2^k - 1$ for some $k \le n_0$, then take $n > n_0$. This proves the desired claim for sufficiently large $n$ and then it suffices to check finitely many cases. 
