Sum of Exponents and Primes Is it necessary that for every prime $p$, there exists at least one of $1+p$, $1+p+p^2$, $1+p+p^2+p^3$, ... that is a prime? Is it also true that an infinite number of $1+p$, $1+p+p^2$, $1+p+p^2+p^3$, etc. are primes?
I am thinking of using something similar to Euclid's proof of the infinitude of primes, given that similar stuff happens. 
Clarifications: My questions are FOR ALL P
 A: For the case as simple as $p=2$, we don't even know the answer to the last question. Primes of this form are called Mersenne primes: http://en.wikipedia.org/wiki/Mersenne_prime.
For the first question note that $1+p+\cdots+p^n=11\cdots 1$ ($n+1$ ones) in base $p$. The primes of this form are called repunit primes. It is known that there are no base $4$ or base $9$ repunits primes. I don't think it is known whether or not there exists a prime number $p$ such that there are no base $p$ repunit primes. See also: http://en.wikipedia.org/wiki/Repunit#Repunit_primes.
EDIT: In fact, there do not exist repunit primes in any quadratic base (except for $1$):
Let $k\in\mathbb{N}-\{1\}$. Then:
$$1+k^2+\cdots+(k^2)^n=\frac{(k^2)^{n+1}-1}{k^2-1}=\frac{(k^{n+1}-1)(k^{n+1}+1)}{k^2-1}$$
Now if $n$ is odd, we have $k^{n+1}=(k^2)^{m}$ (with $m=\frac{n+1}{2}$), hence:
$$k^{n+1}-1=(k^2)^m-1=(k^2-1)\left((k^2)^{m-1}+(k^2)^{m-2}+\cdots+1\right)$$
So then $k^{n+1}-1$ is divisible by $k^2-1$, and hence:
$$\frac{(k^{n+1}-1)(k^{n+1}+1)}{k^2-1}$$
is divisible by $k^{n+1}+1$ (so not prime).
If $n$ is even then:
$$k^{n+1}+1=(k+1)(k^n-k^{n-1}+k^{n-2}-k^{n-3}+\cdots+1)$$
And
$$k^{n+1}-1=(k-1)(k^n+k^{n-1}+\cdots+1)$$
Hence
$$\frac{(k^{n+1}-1)(k^{n+1}+1)}{k^2-1}=\frac{(k^{n+1}-1)(k^{n+1}+1)}{(k-1)(k+1)}$$ $$=(k^n+k^{n-1}+\cdots+1)(k^n-k^{n-1}+k^{n-2}-k^{n-3}+\cdots+1)$$
is divisible by $(k^n+k^{n-1}+\cdots+1)$.
