Solve this inequality with nested radicals (possibly by induction) I tried to solve this problem by induction but didn't succeed. Given the series
$$ a_n = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{... + \sqrt{n}}}}}$$ 
Prove that $a_n < 2 (\forall n \in \mathbb{N^*}) $
Now I thought that maybe I could find a reccurence formula. I haven't found one. Another way I tought of was squaring both sides and substracting the number before the radical n times but that made it more complicated. Can someone lend me a hand on this?
 A: Fix $x \ge 0$, and define $(a_n)$ by
$$
a_n 
=
\sqrt{x^2 + 
 \sqrt{x^4 + 
  \sqrt{x^8 + 
   \sqrt{\cdots + 
    \sqrt{x^{2^n}}
   }
  }
 }
}
$$
Then for all $n > 1$, we have the relation
$$a_n^2=x^2+xa_{n-1}$$
Now let $x = \sqrt{5}-1$.

Claim:$\;a_n < 2$, for all $n$.

Proceed by induction on $n$.

For $n=1$, we have $a_1=\sqrt{x^2}=x = \sqrt{5}-1< 2$.

Suppose $a_n < 2$, for some positive integer $n$.

Then we get
$$a_{n+1}^2 =  x^2+xa_n < x^2+2x = (\sqrt{5}-1)^2+2(\sqrt{5}-1)= 4$$
so $a_{n+1} < 2$, which completes the induction.

Next, compare $x^{2^n}$ and $n$ . . .

Claim:$\;x^{2^n} > n$, for all $n$.

Proceed by induction on $n$.

For $n=1$, we have $x^{2^1} = x^2 = (\sqrt{5}-1)^2 > 1$.

For $n=2$, we have $x^{2^2} = x^4 = (\sqrt{5}-1)^4 > 2$.

Suppose $x^{2^n} > n$, for some positive integer $n \ge 2$.

Then we get
$$x^{2^{n+1}}=\left(x^{2^n}\right)^2 > n^2 \ge 2n > n+1$$
so $x^{2^{n+1}} > n+1$, which completes the induction.

Hence, for all $n$, we have
$$
\sqrt{1 + 
 \sqrt{2 + 
  \sqrt{3 + 
   \sqrt{\cdots + 
    \sqrt{n}
   }
  }
 }
} < 
\sqrt{x^2 + 
 \sqrt{x^4 + 
  \sqrt{x^8 + 
   \sqrt{\cdots + 
    \sqrt{x^{2^n}}
   }
  }
 }
} < 2
$$
A: It is well known that the iteration $x\mapsto \sqrt{1+x}$ with starting point $x=0$ converges to $\varphi=\frac{1+\sqrt{5}}{2}$, i.e.
$$ \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}=\frac{1+\sqrt{5}}{2}.\tag{1} $$
If we multiply both sides by $2^{1/4}$ we get
$$ \sqrt{2^{1/2}+\sqrt{2+\sqrt{4+\sqrt{16+\ldots}}}}=\frac{1+\sqrt{5}}{2^{3/4}}<2\tag{2} $$
and the LHS of $(2)$ is blatantly larger than any $a_n$ since $2^{2^{n-2}}\geq n$ for any $n\geq 1$.
Actually
$$ \lim_{n\to +\infty} a_n = \sup_{n\geq 1}a_n \approx 1.75793\ldots < \frac{2495-\sqrt{5}}{1418}.\tag{3}$$
A: $$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{... + \sqrt{n}}}}}\le \sqrt{1\sqrt{2\sqrt{3\sqrt{...\sqrt{n}}}}} = \prod_{k=1}^nk^{\frac{1}{2^k}} < 1.7$$ 
