Show that $\frac{1}{e^\gamma \text{log }x + O(1)} = \frac{1}{e^\gamma\text{log }x} + O\left(\frac{1}{(\text{log }x)^2}\right)$

Question

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.

Answer 1

Would an answer like this work?:

\begin{align} \frac{1}{e^\gamma\text{log }x + O(1)} &= \frac{1+O(\text{log }x)}{e^\gamma\text{log }x + O((\text{log }x)^2)}\\ &=\frac{1}{e^\gamma\text{log }x}\left(\frac{1+O(\text{log }x)}{1+O(\text{log }x)}\right)\\ &=\frac{1}{e^\gamma\text{log }x}(1+O(\text{log }x)) \end{align}