It is well known that $$ \zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}, \quad \Re[s] > 1, \tag{1}$$ but, if $p \leq N$ denotes the primes less than or equal to $N$, can $(1)$ be used to deduce that $$\sum_{n = 1}^N \frac{1}{n^s} \sim \tilde K \cdot \prod_{p \leq N} \frac{1}{1 - p^{-s}} \quad \text{as} \quad N \to \infty$$ and, in particular, $$\sum_{n = 1}^N\frac{1}{n} \sim K \cdot \prod_{p \leq N} \frac{1}{1 - 1/p} \quad \text{as} \quad N \to \infty, \tag{2}$$ for some constants $\tilde K, K > 0$? Here, $\sim$ denotes asymptotic equality. There is a sharp estimate for $K$, namely $K = e^{-\gamma}$, where $\gamma$ is the Euler-Mascheroni constant (I believe it is one of Mertens' theorems). In any case, does $(2)$ follow from $(1)$?
Edit: Note that I am not looking for a proof of Mertens' theorem. I just wish to show that $K > 0$ exists and $K \ne 1$.