Prove the product of a sum of powers of primes diverges I need to prove that this is divergent:
$$\prod_{\stackrel{p\leq N}{p\text{ prime}}}\left(1 + \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^k} + \cdots\right),$$
where the expression inside of the product is:
$$\frac{1}{(1-\frac{1}{p})} = (1 + \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^k} + \cdots).$$
 A: The fastest way is to use the Fundamental Theorem of Arithmetic (Unique factorization into primes).  You can show that in a combinatorial sense $$\prod_p \left( 1+\frac{1}{p}+\frac{1}{p^2}+\cdots\right)=\sum_{n=1}^\infty \frac{1}{n}.$$  I mean that if the product was expanded, each term on the right would appear exactly once.  This then proves the divergence of your product since the harmonic series diverges.  This is also known as Eulers Proof of the infinitude of the primes.
Hope that helps,
Mertens Estimate:  If you are curious, we can also say things about the rate of divergence of this product.  It is one of Mertens Estimates which says $$\prod_{p\leq N} \left(1-p^{-1}\right)^{-1} = e^\gamma \ln N + O(1)$$ where $\gamma$ is the Euler-Mascheroni Constant.  This means that the product you are looking at grows roughly at the same rate as the logarithm.  This is to be expected from the first part since the divergence of the harmonic series is also like the logarithm, and this argument can be made rigorous.  However the constant requires slightly more work to compute.
Added:  This Wikipedia page is related, and a little interesting too.
A: Hint: Expand the partial product using all primes $p\leq N$ for some large $N$, and compare with the partial sum of the harmonic series.  
A: Hints:  1)inside the parentheses you have a geometric series which can be summed. 2) $\sum_{p\text{ prime}} \frac{1}{p}$ diverges
