¿Existe una identidad general para los infinitos radicales? $\sqrt{n^{0}+\sqrt{n^{1}+\sqrt{n^{2}+\sqrt{n^{3}+...}}}}$

Question

¿Existe una identidad general para los infinitos radicales? $$f(n)=\sqrt{n^{0}+\sqrt{n^{1}+\sqrt{n^{2}+\sqrt{n^{3}+...}}}}$$

For $ n = 0,1,4$ we get $ f (n) = 1, φ, 2$ respectively (If you allow that $ 0 ^ 0 = 1)$ but will there be any other value of $ n$ such that $ f (n) ∈ \ mathbb {Z}$ or even has a closed-form expression? This is actually such an interesting nested radical, that I was unable to even find a good proof for $ f (4) = 2$. Most of them were reverse-engineering the radical and showing but not proving.

So is there a closed form expression for this as general or even for any other value of $ m$ I missed out?

A C++ code for computing this can be given as;

#include <iostream>
#include <math.h>

using namespace std;

long double seq( unsigned long, long double, unsigned long = 0 );

int main() {
    unsigned long m = 0; long double n;
    cout << "Enter integer r: ";
    cin >> m; cout << "Enter n: ";
    cin >> n; cout << seq( m, n); return 0;
}

long double seq( unsigned long m, long double n, unsigned long i) {
    return sqrt( pow(n,i)+(i==m?0:seq(m,n,i+1)) );
}

Where input $ r$ means how many times you need to continue the radicals, and $ n$ will be the number you want to put in.

Edit;

Maybe a trivial finding, but I also got that

$% # PS