$\sqrt[7]{3+\sqrt[7]{3+\sqrt[7]{3+\dotso}}}$ $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}}$$\newcommand{\root}[4]{\sqrt[\leftroot{#2}\uproot{#3}#1]{#4}}$$\newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}$$ \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}$$\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}$

Find the value of $$\root{7}{4}{3}{3+\root{7}{2}{1}{3+\root{7}{1}{2}{3+\cdots}}}$$

What I tried? I apply the same steps on these:
$$x=\root{n}{-1}{1}{t+x}$$
$$x^n=t+x$$
$$\underline{x^n-x-t=0}$$
That was where I was stuck.
Any way to solve that kind of equations?
Or how to solve this:
$$x^7-x-3=0$$
Note: I don't want to use numerical approximation.
NOTE TO READERS: The answers are false. It can be solved like explained here: https://arxiv.org/pdf/math/9411224v1.pdf
 A: Please see this.
You really don't have a choice but to use numerical approximations.
A: These equations cannot generally, I believe, be solved in terms of only a finite number of the basic arithmetical operations, including roots and the exponential and logarithm. Sometimes they can, but there is a lot of complex theory behind this which I have not had enough experience in to give a definitive answer for your particular problem as to this theory's ruling. This is called "Galois theory", and in fact, problems like this were exactly what motivated the eponymous mathematician (who died sadly very young before he could produce even more great stuff) to investigate it. But basically the theory deals with, I believe, how certain operations transform the roots (solutions) - if they transform in the right way, it will have a finite solution in terms of elementary arithmetical operations, if not, it will not. However as I said, I don't have much knowledge in this area.
It is possible though to use other types of operations, for example, if you add hypergeometric functions, you can solve an equation of general form $x^n - x - a = 0$ with a finite hypergeometric representation. But these are defined in terms of infinite sums, so in another sense are perhaps not much different from your infinite radical you began with in terms of "satisfactoriness". (Arguably the infinite radical is already a very elegant-looking expression of the solution to the polynomial in its own right and the hypergeometric series is rather more sticky.) On the other hand, they are extremely useful operations, and can be manipulated in a crazy number of ways, moreso then I believe the infinite radical can be.
A: For sure, since tou face a septic polynomial, only numerical methods will give the answer.
However, we can approximate the solution using $[1,n]$ Padé approximants built around $x=1$ and get the following results
$$\left(
\begin{array}{ccc}
n & x_n & \approx \\
 1 & \frac{13}{11} & 1.18182 \\
 2 & \frac{16}{13} & 1.23077 \\
 3 & \frac{765}{622} & 1.22990 \\
 4 & \frac{3343}{2721} & 1.22859 \\
 5 & \frac{43858}{35695} & 1.22869 \\
 6 & \frac{191739}{156044} & 1.22875 \\
 7 & \frac{419121}{341099} & 1.22874 \\
 8 & \frac{5497012}{4473715} & 1.22874 \\
 9 & \frac{48064227}{39116797} & 1.22874 \\
 10 & \frac{210129473}{171012676} & 1.22874
\end{array}
\right)$$
