Prove by Induction $\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}} = \left[\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right]^{2^{n-1}} $ $a_1=\frac{1}{2}(a_0+\frac{A}{a_0})$;
$a_2=\frac{1}{2}(a_1+\frac{A}{a_1})$; and
$a_{n+1}=\frac{1}{2}(a_n+\frac{A}{a_n})$ for $n \geq 2$; where $a\gt 0$, $A\gt 0$.
Prove:
$$\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}} =
\left[\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right]^{2^{n-1}}
$$
Remarks:(This question i was doing for my exam tomorrow; and i just got stuck)
I have started with applying Second Principle of Mathematical Induction.
I have never done this type of question earlier.
Any help or hint will be appreciated.

Let us start:
\begin{align}
\frac{a_n -\sqrt{A}}{a_n+\sqrt{A}} &\underbrace{=}_{*} 
\frac{\frac{1}{2}a_{n-1}+ \frac{1}{2}\frac{A}{a_{n-1}} -\sqrt{A}}{\frac{1}{2}a_{n-1}+ \frac{1}{2}\frac{A}{a_{n-1}} +\sqrt{A}} \underbrace{=}_{**} 
\frac{\frac{1}{2}a_{n-1}^ 2+ \frac{1}{2}{A} -\sqrt{A}a_{n-1}}{\frac{1}{2}a_{n-1}^ 2+ \frac{1}{2}{A} +\sqrt{A}a_{n-1}} \\
&\underbrace{=}_{***} \frac{a_{n-1}^ 2+ {A} -2\sqrt{A}a_{n-1}}{a_{n-1}^ 2+ {A} +2\sqrt{A}a_{n-1}}
=\frac{(a_{n-1}-\sqrt{A})^2}{(a_{n-1}+\sqrt{A})^2} \\
&=\Bigr(\frac{a_{n-1}-\sqrt{A}}{a_{n-1}+\sqrt{A}}\Bigl)^2\\
&\underbrace{=}_{****}\Bigr(\Bigl(\Bigl(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\Bigl)^2\Bigr)^2 \ldots \Bigr)^2\\
&=  \Bigl(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\Bigl)^{2^{n-1}}
\end{align}
Here, I used the recurrence relation at $*$, multiplied enumerator and denominator with $a_{n-1}$ at $**$ and with 2 at $***$ and used induction at $****$.

Background:
Why is this of any interest?
Well first of all, note that the sequence $a_n$ converges towards $\sqrt{A}$. One possible derivation of the sequence is the application of Newton's method to a function that has a root at $\sqrt{A}$. This could be $f(x) = x^2-A$. Newtons's method then reads:
\begin{align}
x_{k+1} & = x_k -\frac{f(x_k)}{f'(x_k)} = x_k -\frac{x_k^2-A}{2x_k} =\frac{x_k^2+A}{2x_k}\\
&= \frac{1}{2} \Bigl(x_k+\frac{A}{x_k}\Bigr)
\end{align}
Which is exactly your sequence. So if we want to estimate 
$$ \frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}=c$$
where c is the value we derived from the prooved formula, we can use this information to find out, how good we approximate $\sqrt{A}$.
\begin{align}
&\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}=c \\
\Leftrightarrow&a_n-\sqrt{A} = c(a_n+\sqrt{A}) \\
\Leftrightarrow& a_n(1-c) = (1+c)\sqrt{A} \\
\Leftrightarrow& a_n= \frac{1+c}{1-c}\sqrt{A} \\
\end{align}
So just by having a starting value $a_0$, we already know good solution $a_n$ will be, in sense of approximating $\sqrt{A}$. This means, that we can choose a $n$ a priori, for which $|a_n-\sqrt{A}|<\epsilon$, for a given $\epsilon$.
