What is the value of $\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots }}}}$? How can we find the value of
$$\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots}}}}=?$$  
My Approach:
Let $$f(n)=\sqrt[n]{1+ \sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}}} \tag{1},$$
then $f(2)$ is our solution.
So, doing $n$th power in both sides of $(1)$, we get:
$${ \{ f(n) \} }^n =1+f(n+1)$$
$$\implies { \{ f(n) \} }^n - f(n+1) = 1 \tag{2}$$
Now how can I solve $(2)$ ? Any help please…

Estimation of upper bound.
L signs the value of the expression, then we can write: 
\begin{align}L^2=1+\left(1+(1+\cdots+(1+(1+s)^{{1/n}})^{{1/{n-1}}}\cdots)^{1/4}\right)^{1/3}\end{align}
where 
\begin{align}s=\sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}} \tag{1}\end{align}
Accordance with Bernoulli$^,$s inequality if $0\le{r}\le1$
and $x\ge-1$ then $(1+x)^r\le(1+xr)$ 
In our case $r=\frac{1}{k}, k=3,4, ....n $ and $s\gt1  $.
First apply to $(1+s)^{1/n}$ we have that  $\le1+\frac{s}{n}$ 
then outward apply to all power factor we get the following: 
\begin{align}L^2\le1+1+\left(1+(1+\cdots+(1+(1+\frac{s}{n}))\frac{1}{n-1}\cdots)\frac{1}{4}\right)\frac{1}{3}\end{align}
Perform the multiplications: 
\begin{align}L^2\le2 +2(\frac{1}{3!} +\frac{1}{4!}+......\frac{1}{(n-1)!}+\frac{s}{n!})\end{align}
Namely
\begin{align}L^2\le2 +2\sum_{k=3}^n\frac{1}{k!}\end{align}
We could apply to $s$ the above procedure then we have: 
\begin{align}L^2\le2 +2\sum_{k=0}^{\infty}\frac{1}{k!}-5\end{align}
\begin{align}L\le \sqrt{2e-3}=1,56094\end{align}
