How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?

We have

$$\sqrt{2}=2^{\frac{1}{2}}.$$ Exciting, no? We also have $$\sqrt{2\sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}},$$ and $$\sqrt{2\sqrt{2\sqrt{2}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}.$$ For the next term, we take the previous term, multiply by $2$, getting exponent of $2$ equal to $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, then take the square root, getting exponent of $2$ equal to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$. The pattern continues, since the "next term" is always obtained by the same process.

It is well-known that the series $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ converges to $1$. Since the function $f(x)=2^x$ is continuous, our original sequence converges to $2^1$.

Put

$$a_n:=\underbrace{\sqrt{2\sqrt{2...\sqrt 2}}}_{n\text{ square roots}}\Longrightarrow a_n=\sqrt{2a_{n-1}}$$

Since you've already proved the sequence $\,\{a_n\}\,$ converges, assume its limit is $\,x\,$ , so

$$x=\lim_{n\to\infty}a_n=\lim\sqrt{2a_{n-1}}=\sqrt{2x}\Longrightarrow x^2=2x$$

and since it's trivial that $\,x\neq 0\,$ then you can cancel above and get what you want.