Proof: $2^{2^t}-1$ is divisible by at least $t$ distinct primes A question into the elementary number theory. Proof: $2^{2^{t}}-1$ is divisible by at least $t$ distinct primes.  
My ideas about the issue are the following:
Distinct primes, call them: $p_{1};p_{2};...;p_{k}$, we want to show that $k>t$ or $k=t$
So: $2^{2^{t}}=1$ (mod $p_{i}$) for all $1<i<k$ which perhaps could be solved by the Chinese remainder theorem. (and $i=1$, $i=k$)
And we know that $2^t\ |\ \phi(p_{i})=p_{i}-1$ so $2^t=1$ mod($p_{i}$) for all $1<i<k$ and $i=1$; $i=k$. 
These are just observations, without a proof-strategy. Can anybody give me some direction? Thanx in advance!
 A: Hint
We have the following:
$$
2^{2^t}-1= (2^{2^{t-1}}-1)(2^{2^{t-1}}+1)$$
for $t\geq 1$. 
A: Meanwhile, the question has been solved.
You can split up:
$(2^{2^{t}}-1)$
$=(2^{2^{t-1}}-1)(2^{2^{t-1}}+1)$
$=(2^{2^{t-1}}+1)(2^{2^{t-2}}-1)(2^{2^{t-2}}+1)=...=(2^{2^{t-1}}+1)(2^{2^{t-2}}+1)\cdot{}...\cdot{}(2^{2^{t-t}}+1)(2^{2^{t-t}}-1)$
The last factor is equal to 1, so $(2^{2^{t}}-1)$ gives us a product of $t$ different brackets. Now we want to show that $(2^{2^{t}}-1)$ brings $t$ different prime numbers. So we have to investigate if $(2^{2^{t-i}}-1)$ is prime for all $1<i<t$ and $i=1,i=t$.
Using induction:
•$t=1$: $(2^{2^{1}}-1)=3$, so there are $t=1$ different prime numbers. Correct.
Supose it is true for $t=k$, there are $t=k$ different prime numbers.
•$(2^{2^{k}}-1)$ brings $t=k$ different a prime numbers.
Investigate how many prime numbers $(2^{2^{t}}-1)$ produces for $t=k+1$.
•$gcd((2^{2^{t}}-1);(2^{2^{t}}+1))=1,2$, but it could not be 2 because $(2^{2^{t}}-1)$ is odd, so it follows that the $gcd$ is one, and so brings a new prime number.
Therefore, $(2^{2^{t}}-1)$ is divisible by at least $t$ different prime numbers. :-)
