Evaluate $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$ $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
My Attempt: I tried to use the regular way.
$A=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
$A^2=1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}$
But then I saw that nothing happened twice and I stopped. Any hints?
It's better to don't use limits but if you used no problem. If there is an answer that doesn't use limits will accept.
update1: The question that look likes this  wants to find the limit but I want a much easier way to solve it. But if there isn't any answer easier answer the limition one will accept.
update2: You solve the problem when you know the answer is 3 think that you don't know that the answer is 3 then solve.

Setting the Scene

Evaluate $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times  }}$

Let $$f_n(0)=\sqrt{1+n}$$also that $$f_n(k)=\sqrt{1+(n-k)f_n(k-1)}.$$
Then clearly $0<f_n(0)<n+1$ when $n>0$. Now if we assume that $f_n(k)<n+1-k$ and by induction that we see that
$$
f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1)
$$
for all k. The expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$ and thus converges. Hence

$\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times  }}=3$

EDIT 
Showing that $3$ is the limit can be achieved by writing
\begin{align}
3 &= \sqrt{9} \\
&= \sqrt{1 + 8}\\
 &= \sqrt{1 + 2 \cdot 4} \\
&= \sqrt{1 + 2\sqrt{16}}\\ 
&= \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}\\
& = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} \\
&= \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}\\
&= \ldots
\end{align}
and so on.
