We know that : $ \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p} \frac{1}{1 - p^{-s}} $ From now on, assume the sum over $ n $ means $1$ to infinity. And assume that the sum or product over p means over the primes, i.e. $2, 3, 5, 7,... $
Therefore, for $ s= 1 $ we have $\sum_{n} \frac{1}{n} = \prod_{p} \frac{1}{1 - \frac{1}{p}} $
Now we will abbreviate the harmonic series as $S_h$. Note that this series diverges. Next we will take the log of both sides.
$\implies \log(S_h) = \log(\prod_p \frac{1}{1 - \frac{1}{p}}) = \sum_p \log(\frac{1}{1 - \frac{1}{p}}) = -\sum_p \log(1 - \frac{1}{p}) $
Now we will use the Taylor series $\log(1-x) = -\sum_n \frac{x^n}{n} $ with $x = \frac{1}{p}$ to get:
$\log(S_h) = -\sum_p\bigg[-\sum_n \frac{1}{np^n} \bigg] = \sum_p \bigg[\sum_n \frac{1}{np^n} \bigg] < \sum_p \bigg[\sum_n \frac{1}{p^n} \bigg]$
Now use the formula for the geometric series starting at $n=1$, i.e. $\frac{x}{1-x}$.
$\implies \log(S_h) < \sum_p \bigg[\frac{1/p}{1 - \frac{1}{p}} \bigg] = \sum_p \frac{1}{p-1} $
And since $p>2 \implies p - 2> 0 \implies 2p -2 > p \implies p - 1> \frac{p}{2} \implies \frac{1}{p-1} < \frac{2}{p}$ we have:
$\log(S_h) < \sum_p \frac{1}{p-1} < \sum_p \frac{2}{p} $
Since $S_h$ diverges, then so does $\sum_p \frac{1}{p} $ . $\square$
I'm worried about the way I treated $S_h$ (the harmonic series), since it diverges and is not a number. I'm doing normal calculus operations on something that is not a number. But each individual step makes sense to me, so I don't know.