Can the result for $\prod_{p\ge 3}\left(1-\frac 2p\right)$ be extended for general $\prod_{p\ge n+1}\left(1-\frac np\right)$? In the excellent answer at , the identity $\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\log\left(1- \frac{1}{(p-1)^2} \right)=\log\left(1- \frac{2}{p} \right)$ was put forward as a starting point.  I thought I would try a similar thing for estimating $\prod_{3\lt p\le x}\left(1-\frac 3p\right)$ and thus use some identity which might be along the lines of  $$\log\left(1-\frac 3p\right)=\log\left(1-\frac 1p\right)+\log\left(1-\frac 2p\right)+\log\left(1-\frac 2{p-1}-\frac 2{p-2}\right)$$  but as I work with these quantities I realize that no identity will completely remove the factor $p-3$ from the results entirely.  In my search results I keep finding the same "Euler Product" and "Mertens' Theorem" pages on Wikipedia and Mathworld, but none of those pages has a $p-3$ factor listed that I have noticed, nor any suggestion as to how to approach the broader question.  I am not especially worried about the accuracy of the error term, as long as the error is not asymptotically as big as the rest of the result.  Are there known results that my primitive searches haven't found yet?  Are there any reference materials I could look at that might help me work out some of these bounds myself, or perhaps some simple observation that has a relatively large error term but simplifies the approach in general?
Derivation of a generalization of Mertens' Third Theorem. One of Mertens' Theorems states $$\prod_{p\le x}(1-\frac{1}{p})\sim \frac{e^{-\gamma}}{\ln(x)}.$$ I have seen a generalized version that states  $$\prod_{m&lt;p\le x}(1-\frac{m}{p})\sim \frac{c(m)}{(\ln(x))^m},$$ where $m$ is a positive integer and $c(m)$ is a real number that depends on $m$. How can I derive this result? Is there a way to find what the values of $c(m)$ are? Thank you very much.
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