Show that $\frac{1}{e^\gamma \text{log }x + O(1)} = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\text{log }x)^2}\right)$ Show that $\frac{1}{e^\gamma \text{log }x + O(1)} = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\text{log }x)^2}\right)$
I'm using one of Merten's estimates in a proof, the one that states 
\begin{align}
\prod\limits_{p \leq x}\left(1-\frac{1}{p}\right)^{-1}=e^\gamma\text{log }x +O(1)
\end{align}
To find $\prod\limits_{p \leq x}\left(1-\frac{1}{p}\right) = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\operatorname{log}x)^2}\right)$
 but I am loss at how to show this, any help is appreciated greatly. 

You might want to work with logarithms. Since $\log x+\mathcal O(1)=\log x\{1+\mathcal O(1/\log x)\}$, we have
$$
-\log\prod_{p\le x}\left(1-\frac1p\right)=\gamma+\log\log x+\mathcal O\left(1\over\log x\right)
$$
Now multiply both sides by $-1$ and exponentiate, so we get
$$
\prod_{p\le x}\left(1-\frac1p\right)={e^{-\gamma}\over\log x}\left\{1+\mathcal O\left(1\over\log x\right)\right\}
$$
Finally, expanding the parenthesis, we get the desired result.
