# How to derive the exponential Lorentz factor approximation?

In the book "Essential Relativity: Special, General and Cosmological" by W. Rindler, the author uses the approximation

$$\gamma\approx10^n,$$

where $\gamma$ is the Lorentz factor and, $\frac{v}{c}=0.99...95$ (with $2n$ nines). I am finding it very hard to derive this approximation, and the only references that I can find online are provided by Rindler without derivation (http://www.scholarpedia.org/article/Special_relativity:_kinematics, near Figure 2).

The book mentions that you can use the fact that $1-\frac{v^2}{c^2}=(1+\frac{v}{c})(1-\frac{v}{c})$ to help in the derivation, so that $$\gamma=(1+\frac{v}{c})^{-\frac{1}{2}}\cdot(1-\frac{v}{c})^{-\frac{1}{2}}.$$ This seems to lend itself quite nicely to a binomial expansion, which I am confident in doing, but it still eludes me how to link this to $10^n$. This seems to be a very useful approximation when $v\rightarrow c$ but I'd love to be able to understand the reasoning behind it. Any hints towards a solution would be greatly appreciated.

The trick is to write $\beta = v/c = 0.\underbrace{99\ldots9}_{2n\text{ 9's}}5$ as $\beta = 1 - 5\times 10^{-(2n+1)}$. Then you have an expression to plug in for $1 - \beta$.
If you want to use the factorization the book described, you can approximate $1 + \beta \approx 2$. Alternatively, you can expand the definition $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ in a series around $1 - \beta = 0$. Substituting $x = 1-\beta$ may clarify the process.