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?

I think that OP's guess was okay, here's a different approach.
Let $a_1 = 3$ and let $a_{n+1} = 3 + \sqrt{3a_n}$ for $n\geq 1$. Then the first few elements are
\begin{align*}
a_1 &= 3,\\
a_2 &= 3+ \sqrt{3^2},\\
a_3 &= 3+\sqrt{3^2 + \sqrt{3^4}},\\
a_4 &= 3+ \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8}}},
\end{align*}
so it is reasonable to define
\begin{align*}
3+ \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8+\ldots}}} := \lim\limits_{n\to\infty} a_n.
\end{align*}
We first need to make sure that the limit exists. We will use the theorem saying that any monotone and bounded sequence is convergent, it is often useful for recursive sequences such as $a_n$.
Monotonicity: we will prove it by induction. First note that $a_2 \geq a_1$. Now assume that for some $n\geq 1$ we have $a_{n+1}\geq a_n$. Then
\begin{align*}
a_{n+2} - a_{n+1} = 3 + \sqrt{3a_{n+1}} - (3 + \sqrt{3a_{n}}) = \sqrt{3}(\sqrt{a_{n+1}} - \sqrt{a_n}) \geq 0,
\end{align*}
where the last inequality follows from the induction hypothesis and the fact that the square root is increasing. Thus, by induction we get that $a_{n+1}\geq a_n$ for all $n\geq 1$, so the sequence is increasing.
Boundedness: obviously the sequence is bounded from below by 0. To find a reasonable upper bound let us first compute what we think the limit should be (if a sequence is increasing, then its limit is also its upper bound). Note that
\begin{align*}
\lim\limits_{n\to \infty} a_{n} = \lim\limits_{n\to \infty} a_{n+1} = 3 +3\lim\limits_{n\to\infty}\sqrt{a_n},
\end{align*}
so if we let $x = \lim\limits_{n\to \infty} a_{n}$, we get that $x = \frac 32(3 + \sqrt{5}).$
To make the computations easier, we will show that $a_n \leq 27$ by induction ($27\geq x$ and it works nicely if we plug it into $\sqrt{3\cdot a_n}$). It is of course true for $a_1$. If $a_n\leq 27$ , then
\begin{align*}
a_{n+1} = 3 + \sqrt{3a_n} \leq 3 + \sqrt{3\cdot 27} = 3+9 = 12 \leq 27.
\end{align*}
Thus we get that $a_n$ is bounded and increases, so it has a finite limit. By the above computation, the limit is equal to $x$.
