Prime factorization of $10^n+1$ I was just playing with these numbers and it seems to me that the numbers of the form $10^n+1$, where $n>2$ are composite. I can prove that $10^n+1$ can't be prime unless $n$ is a power of $2$, but I do not know for which values of $n$, these numbers are prime, except for $n=2$. Is it an open problem or is there any result on this? Thank you in advance.
 A: $10^{2^n}+1$ is a generalized Fermat prime $F_{n}(10)$.
Landau's fourth problem asks if there are infinitely many generalized Fermat primes $F_n(a)$:

As of 2013, all four problems [including yours] are unresolved.

Here are some more numbers, till $n=20$. 
A: Updated to add: Due to a typo in the original question, it turns out that this proof is already known to the OP. Never mind, I'll leave it here anyway.
We can write $n=2^km$ where $m$ is odd. Then $10^n \equiv (-1)^m \equiv -1$ mod $(10^{2^k}+1)$, so $10^n+1 \equiv 0$ mod $(10^{2^k}+1)$. Thus $10^n+1$ is composite unless $n=2^k$, i.e. $n$ is a power of $2$.
For instance:


*

*if $n$ is odd, then $10^n+1 \equiv (-1)^n+1 \equiv 0$ mod $11$  

*if $n \equiv 2$ mod $4$, put $n=4m+2$, then $10^n+1 \equiv 100^{2m+1}+1 \equiv (-1)^{2m+1}+1 \equiv 0$ mod $101$  

*similarly, if $n \equiv 4$ mod $8$, then $10^n+1 \equiv 0$ mod $10001$  

*if $n \equiv 8$ mod $16$, then $10^n+1 \equiv 0$ mod $100000001$  


and so on.
If $n$ is a power of $2$, you claim that you can prove that $10^n+1$ is prime, but that's not right. In fact $10^n+1$ is prime for $k=1$ (with $10^n+1 = 101$) and composite for $k=0,2,3,4,5$. I can see no reason to rule out prime values for higher values of $k$.
