Solving $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$ 
How to find the value of $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$

I tried to solve it and found a relation that if I assume the given expression to be something say $x$, then following result holds true.
$$\boxed{x = 3 + \sqrt{3x}}\hspace{4cm}\bf ...(1.)$$
Can be proved as,
$$x = 3 + \sqrt{3(3 + \sqrt{3x})}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3x}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3(3 + \sqrt{3x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7(3 + \sqrt{3x})})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{15}x}}}}$$
If I continue this process repeatedly,
$$x= 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^{8} + \sqrt{3^{16} +\sqrt{...}}}}}$$
But how can I make sure that $(1.)$ is absolutely correct? Also how can I make sure that "If I continue this process repeatedly" I would get the same expression?

If we have
$$a_{n+1}=x \sqrt{x\,a_n} \qquad \text{with}\qquad a_0=x$$
$$a_n=x^{b_n} \implies  x^{b_{n+1}}=x \sqrt{x^{b_n+1}} \implies b_{n+1}=\frac 12 b_n+\frac 32$$
$$b_n=c_n+3\implies c_{n+1}=\frac 12 c_n$$
Finishing and back to $a_n$
$$a_n=x^{3-2^{1-n}}\quad \to \quad x^3$$
