showing that $\log(N) \leq \prod_{n \leq N} {(1-p^{-1})^{-1}}$ i can't see that $H_n \leq \prod_{n \leq N}{(1-p^{-1})^{-1}}$
and i can't see that $\log(N) \leq \prod_{n \leq N} {(1-p^{-1})^{-1}}$
p is prime and $H_n$ is harmonic series 
 A: First of all,
since $1/(1-x) = \sum_{k=0}^{\infty} x^k$
if $0 < x < 1$,
for any prime $p$
$1/(1-1/p) = \sum_{k=0}^{\infty} 1/p^k$.
Therefore
$\prod_{p \le n} 1/(1-1/p)
=\prod_{p \le n} \sum_{k=0}^{\infty} 1/p^k
$.
Consider what that last product of sums means:
By unique factorization,
every number with all its prime factors
$\le n$ is represented. 
Written out,
$\prod_{p \le n} \sum_{k=0}^{\infty} 1/p^k
= \sum_{p|m => p\le n} 1/m
$.
This is probably the hardest step to understand,
but it is a well-known number theoretic argument
dating at least back to Euler.
But certainly
all integers $\le n$ have all their prime factors
$\le n$.
(Of course a lot more do also.)
Therefore
$\sum_{p|m => p\le n} 1/m
> \sum_{m=1}^n 1/m
= H_n
$.
(Added a little later.)
To show that
$H_n > \ln n$,
use
$\int_1^n dx/x = \ln n$
and
$1/k > \int_k^{k+1} dx/x$.
A: Take any integer $k\le n$.
Then $k$ can be written as a product of primes $p\le n$.
$$\sum_{k\le n}\frac{1}{k}\le\prod_{p\le n}\left(1+\frac{1}{p}+
\frac{1}{p^2}+\cdots\right)$$The above is a "$\le$" and not a "$=$" because there are many numbers greater than $n$ which can be factored by primes $p\le n$.
Hence the Right side weighs more than the left. For the next part proving $\log(n)<H_n$, as above, suffices. 
