<s>[INST] 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…
 [/INST] Comment, not an answer
We have $$\begin{align}f(2)=\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots}}}}=\left(1+\left(1+(1+\cdots)^{1/4}\right)^{1/3}\right)^{1/2}\end{align}$$ From the Binomial Theorem, $$\begin{align}(1+x)^{1/n}&=1+\frac1nx-\frac{n-1}{n\cdot2n}x^2+\frac{(n-1)(2n-1)}{n\cdot2n\cdot3n}x^3-\cdots\\&=1+\sum_{m=1}^\infty \left(x^m\prod_{k=0}^m\frac{(-1)^{k+1}(kn-1)}{(k+1)n}\right)\end{align}$$ so $$f(2)=1+\sum_{m=1}^\infty \left(\left(1+\sum_{m=1}^\infty \left(\left(\cdots\right)^m\prod_{k=0}^m\frac{(-1)^{k+1}(3k-1)}{3(k+1)}\right)\right)^m\prod_{k=0}^m\frac{(-1)^{k+1}(2k-1)}{2(k+1)}\right)$$
which is essentially an infinite nest of iterations of simple summations and products.
There may be techniques to evaluate this, but none of which I am aware of.
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