Let $n$ be the integer. Prove that $$\sqrt{2\sqrt{4\sqrt{8....\sqrt{2^n}}}} \leqslant n+1$$
SOURCE: BANGLADESH MATH OLYMPIAD
I am a new beginner at the infinite radical and sequence. I don't know about its basic conception.
My Attempt:
After seeing that problem, out of curiosity, I thought that I should calculate the multiplication of left side term.
Than I got something like this
$2^{\frac{1}{2}}$ × $2^{\frac{1}{2}}$ × $2^{\frac{3}{8}}$... × $2^{\frac{k}{2n}} = 2^{\left(\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+...+\frac{k}{2n}\right)}$, where I denoted $2n$ = $2^\text{k}$.
But here I got stuck. The above sequence doesn't follow the pattern. Than how would I get the summation of the sequence?
Moreover, how should I approach to prove the above condition for the integer value of $n$? Any kind of help or clue will be greatly appreciated by this novice and new beginner. Thanks in advance.