Let's play with Exponent tower

Yes, it's true. Let's suppose that

$\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}} }}} } }<2\; \;\;\;\;\;\; (1)$

Squaring both sides:

$2^{\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}} }}} }<4=2^2$

By monotonicity:

$\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}} }} }<2$

Which is similar to $(1)$ , with one square root less. It's now clear that iterating the process we have to end up with

$\sqrt{2} <2$ .

which is true, therefore $(1)$ is true too.

I don't know exactly what @Naren Bhandari wanted to say with "generalize", but I assume it's that: let's consider the sequence

$\begin{array}{c}aa_1=\sqrt{x} \\ a_n=\sqrt{x^{a_{n-1}}} \end{array}$

We can write $a_n$ in this way:

$a_n=\sqrt{x^{a_{n-1}}}=\sqrt{x}^{a_{n-1}}$

Therefore $a_2=\sqrt{x}^{\sqrt{x}}, \; a_3=\sqrt{x}^{\sqrt{x}^{\sqrt{x}} }$ and so on; hence $a_n$ is $\sqrt{x} \uparrow \uparrow n$ using Knuth's up-arrow notation.

For a more formal, although trivial, proof:

We can prove that by induction. The base case is obvious. The induction step too:

$a_{n+1}=\sqrt{x}^{a_n}=\sqrt{x}^{\sqrt{x} \uparrow \uparrow n}=\sqrt{x}\uparrow(\sqrt{x}\uparrow\uparrow n)=\sqrt{x} \uparrow\uparrow (n+1)$

Rewrite $\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}}}}}}}$ as $\left(2^{\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}}}}}}\right)^{\frac{1}{2}} = \left(\sqrt{2}\right)^{\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}}}}}}$ .

For equivalence to hold, $\sqrt{2^{\sqrt{2^{\sqrt{2^{\sqrt{2}}}}}}} = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}$ . Continue the pattern and finally you will get that $\sqrt{2}=\sqrt{2}$ which is obviously true. All that's left is proving that they're both less than $2$ . We know that $\sqrt{2}^2 = 2$ so $\sqrt{2}^{\sqrt{2}} < 2$ and continue this reasoning to prove that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}} < 2$ and hence both are less than $2$ .

For x > 0, the answer will always be closer 0

$\sqrt{a^b} = \sqrt{a}^b$ for $a, b >0$ , since both are positive and their squares are equal. The conclusion follows.