Show that the sequence $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$ converges and find its limit.

The geometric series $\sum_{k=1}^{\infty}(1/2)^k = 1$ .

Hence $x_n = 2^{\sum_{k=1}^n (1/2)^k} \le 2$. Hence, The sequence is bounded above by $2$ and thus the series converges as it is increasing. Now you can use the hint given by ferson2020 to see that the limit $L$ has the property $L = \sqrt{2L}$ and hence $L = 2$.

Added: I am not sure if you proved the formula

$$x_n = 2^{\sum_{k=1}^n (1/2)^k}$$ so I give a proof here. We proceed with induction on n. The base case $n = 1$ holds by definition. Now assume it holds for $x_n$, we show that it is holds for $x_{n+1}$.

We have $x_n = \sqrt{2x_n}$.

So by the inductive hypothesis: $$x_{n+1} = \sqrt{2 \cdot 2^{\sum_{k=1}^n (1/2)^k}} = 2^{({\sum_{k=1}^n (1/2)^k} +1)/2}$$

Now ,

$ S = \sum_{k=1}^n (1/2)^k = 1/2 + 1/4 + \ldots 1/2^n$

and if we multiply this with $1/2$ we get $1/4 + 1/8 + \ldots 1/2^{n+1}$.

So $$S/2 = \sum_{k=2}^{n+1}(1/2)^k$$

Then $$S/2 + 1/2 = \sum_{k=1}^{n+1}(1/2)^k$$

So we have $x_{n+1} = 2^{(S +1)/2}= 2^{\sum_{k=1}^{n+1}(1/2)^k}$ this completes the proof.

If you already proved the exponential form of $x_n$, then, it is easy, as $t\mapsto 2^t$ is continuous, and the exponent $$\sum_{k=1}^n \frac1{2^k} \ = 1-\frac1{2^{n+1}} \ \ \overset{n\to\infty}\longrightarrow \ 1 \ ,$$ so your limit is $2^1=2$.

We have the recursive definition $a_1=\sqrt{2},a_{n+1}=\sqrt{2a_n}.$ (Hopefully, that's clear.)

You've shown that the $a_n$ form an increasing sequence, and you know that $a_1<2.$ For the induction step, we want to show that if $a_n<2,$ then $a_{n+1}<2.$ Note that all the $a_n$ are positive (since $a_1$ is and the sequence is increasing), and note that for $0\leq x<y$ we have $\sqrt{x}<\sqrt{y}$. Thus, if $a_n<2$, then $2a_n<4$, and so $$a_{n+1}=\sqrt{2a_n}<\sqrt{4}=2,$$ as desired.

In general, if you're trying to prove something like this, you'll want to be able to rewrite $a_{n+1}$ in terms of $a_n$ to allow the induction step to work.