why Pi is transcendental number if it will satisfy the algebraic equation as "n" tends to infinity? why Pi is transcendental number if $\pi$ also have algebraic equation like below  which have root at  $x =\pi/3$   as $n$ tends to infinity.
$$\Biggl(\biggl(\Bigl(\bigl((x^2-2^{(n2^1+1)})^2-2^{(n2^2+1)}\bigr)^2-2^{(n2^3+1)}\Bigr)^2...-2^{(n2^{(n-1)}+1)}\biggr)^2-2^{n2^{n}+1}\Biggr)^2 = 3*2^{n2^{(n+1)}}$$ 
For $n = 1$ equation will be $$(x^2-2^3)^2 = 3*2^4$$
and its root  $x_1$ = 1.03527618041. 
For $n = 2$ equation will be $$((x^2-2^5)^2-2^9)^2 = 3*2^{16}$$
and its root is $x_2$ =  1.04420953776041 
and as we increase the value of n the root tends to $\pi/3$.
For n  = 12 ,$3x_{12}$ will be  3.14159264 which is some what close to $\pi$.
Another simplified version of this equation is $${\Biggl(\biggl(\Bigl(\bigl(\frac{\pi}{3*2^n}\bigr)^2-2\Bigr)^2-2\biggr)^2...-2\Biggr)^2} = 3$$
Here n is the same as number of single powered two in above equation. 
Solution of the above equation will be $$\pi = 3*\Biggl(\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{3}}}}}}\Biggr)*2^n$$
and $$n =\text{ number  of $2$ inside  largest  square  root}$$
So if $\pi$ can also be represented as a root of algebraic equation then why $\pi$ is transcendental .
 A: Algebra for the most part is not concerned with sequences and limits. A real or complex number is said to be algebraic if it is a root of a single polynomial equation of finite degree with integer coefficients. This is something that $\pi$ is not.
If you are willing to use limits, then every number is "algebraic" : you can always find a sequence of polynomial equations with integer coefficients whose roots converge to any number. For real numbers, the polynomials can even be linear! So in a sense this is not interesting. The specific equations you outline are interesting, but the conclusion is not as it is true for any number. 
A: All you've shown is that $\pi$ can be approximated by algebraic numbers. But that's not the same as being an algebraic number! When you write 

$\pi$ can also be represented as a root of algebraic equation

you're misusing the term "algebraic equation." There is a sense in which $\pi$ can be thought of as the solution to an "infinite algebraic equation," but that's not the same thing. 
Indeed, the same reasoning would also imply that $\sqrt{2}$ is rational since we can get better and better rational approximations to it (this is expanding on Rijul Saini's comment above). Or that the infinite sum $\sum_{i=1}^\infty 1=1+1+1+1+ ...$ is finite since for each $n$, the partial sum $\sum_{i=1}^n1$ is finite.
In general, you cannot conflate finite objects with their infinite analogues.

This is not to say that there's nothing interesting here. While every real number is "limit algebraic," things get more interesting when we look at functions - thinking about "infinite-degree polynomials" leads to the concept of analytic functions, which is absolutely fundamental. But that's a separate issue.
