Dividing by $\sqrt n$ Why is the following equality true? 
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$

I find this easier to visualize if I write this problem in terms of powers:
$$\dfrac{n^{1/2}}{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}\text{.} $$
Division by $n^{1/2}$ for both the numerator and denominator turns the numerator into $1$ and the denominator into
$$\begin{align*}
\dfrac{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}{n^{1/2}} &= \left[\dfrac{n + \left(n + n^{1/2}\right)^{1/2}}{n}\right]^{1/2}\quad \text{ since }\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^{c} \\
&= \left[\dfrac{n}{n} + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{n}\right]^{1/2} \quad \text{ since }\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \\
&= \left[1 + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{n}\right]^{1/2} \\
&= \left[1 + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{\left(n^{2}\right)^{1/2}}\right]^{1/2} \\
&= \left[1 + \left(\dfrac{n+n^{1/2}}{n^{2}}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^{c} \\
&= \left[1 + \left(\dfrac{n}{n^2}+\dfrac{n^{1/2}}{n^2}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \\
&= \left[1 + \left(\dfrac{1}{n}+\dfrac{1}{n^{3/2}}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a^b}{a^c} = a^{b-c} \\
&= \left\{1 + \left[\dfrac{1}{n}+\dfrac{1^{1/2}}{(n^{3})^{1/2}}\right]^{1/2}\right\}^{1/2} \\
&= \left\{1 + \left[\dfrac{1}{n}+\left(\dfrac{1}{n^3}\right)^{1/2}\right]^{1/2}\right\}^{1/2} \\
&= \sqrt{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}}}}\text{.}
\end{align*}$$
