Consider the set $ S_n$ of all the $ 2^n$ numbers of the type $ 2\pm \sqrt{2 \pm \sqrt {2 \pm ...}},$ where number $ 2$ appears $ n\plus{}1$ times. 
$ (a)$ Show that all members of $ S_n$ are real.
$ (b)$ Find the product $ P_n$ of the elements of $ S_n$.
1. **(a) Show that all members of \( S_n \) are real.**

   We will use mathematical induction to show that all members of \( S_n \) are real.

   **Base Case:**
   For \( n = 1 \), the set \( S_1 \) consists of the numbers \( 2 \pm \sqrt{2} \). Clearly, both \( 2 + \sqrt{2} \) and \( 2 - \sqrt{2} \) are real numbers since \( \sqrt{2} \) is real.

   **Inductive Step:**
   Assume that for some \( k \geq 1 \), all members of \( S_k \) are real and lie in the interval \( (0, 4) \). That is, \( 0 < T(k) < 4 \) for any \( T(k) \in S_k \).

   We need to show that all members of \( S_{k+1} \) are real. Consider any member of \( S_{k+1} \), which is of the form \( 2 \pm \sqrt{T(k)} \) where \( T(k) \in S_k \).

   Since \( 0 < T(k) < 4 \), we have \( 0 < \sqrt{T(k)} < 2 \). Therefore, the expressions \( 2 + \sqrt{T(k)} \) and \( 2 - \sqrt{T(k)} \) are both real and lie in the interval \( (0, 4) \).

   Hence, \( 0 < 2 \pm \sqrt{T(k)} < 4 \), which means all members of \( S_{k+1} \) are real.

   By the principle of mathematical induction, all members of \( S_n \) are real for all \( n \).

2. **(b) Find the product \( P_n \) of the elements of \( S_n \).**

   The elements of \( S_n \) are of the form \( 2 \pm \sqrt{2 \pm \sqrt{2 \pm \ldots}} \), where the number 2 appears \( n+1 \) times. These elements are the roots of the polynomial:
   \[
   ((\ldots((x - 2)^2 - 2)^2 - 2)^2 - \ldots - 2)^2 - 2 = 0
   \]
   where \( -2 \) appears \( n+1 \) times.

   To find the product of the roots of this polynomial, we use the fact that for a polynomial of the form \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \), the product of the roots (taken with their multiplicities) is given by \( (-1)^n \frac{a_0}{a_n} \).

   In our case, the polynomial is constructed recursively, and the constant term \( a_0 \) is 2, while the leading coefficient \( a_n \) is 1. Therefore, the product of the roots is:
   \[
   P_n = 2
   \]

The final answer is \( \boxed{2} \)