real analysis - proof: convergence of recursive sequence (Assignment)

Question: Recursively define a sequence by $x_1=1$ and $x_{n+1}=(\sqrt2)^{x_n}$.

Prove that the sequence $\{x_n\}_{n=1}^{\infty}$ converges.

Attempt: To prove its convergence, I have to show the sequence is bounded and monotone.

I can prove the sequence $x_n\ge 1$ by induction.

I can prove the sequence is monotone increasing $x_n\le x_{n+1}$ by induction.

Since it is monotone increasing, I need to show the sequence is bounded above, but I don't know how to find this.

Could you give some idea? By the way, it is an assignment question.

Answer

Note that $a^b \le a^c$ for $1\le b \le c$ and $a\ge 1$.

$|x_{2}|=(\sqrt{2})^1 \le (\sqrt{2})^2=2$.

$|x_{3}|=(\sqrt{2})^{x_2} \le (\sqrt{2})^{|x_2|} \le (\sqrt{2})^2=2$.

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$|x_{n}| \le 2$.

I'm only just restating what User L KM wrote in their answer.