How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$ $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.

Idea: You can unwrap like this:
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$
if
$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<2^2-1$$
if
$$\sqrt{3+\cdots+\sqrt{2006}}<(2^2-1)^2-2$$
and so on, so we want to show
$$2006 < (((2^2-1)^2-2)^2-\cdots)^2-2005$$

might as well prove it by induction for all $n$ rather than just 2006, so we need to show that
$$n+1 < (((2^2-1)^2-2)^2-\cdots)^2-n$$
implies
$$n+2 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2-(n+1)$$
but thats just
$$2n+3 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2$$
which holds since
$$2n+3 < (n+1)^2$$
for all $n>1$.
