What is a proof of this limit of this nested radical?

For any $2 \le n \le m$, let $\phi_{n,m}(x) = \sqrt[n]{x + \sqrt[n+1]{x + \sqrt[n+2]{x + \cdots \sqrt[m]{x}}}}$. I will interpret the expression we have as following limit.

$$\sqrt{x + \sqrt[3]{x + \sqrt[4]{x + \cdots }}}\; = \phi_{2,\infty}(x) \stackrel{def}{=}\;\lim_{m\to\infty} \phi_{2,m}(x)$$

For any $x \in (0,1)$, we have $\lim\limits_{m\to\infty}(1-x)^m = 0$. This implies the existence of an $N$ so that for all $m > N$, we have

$$(1-x)^m < x \implies 1 - x < \sqrt[m]{x} \implies \phi_{m-1,m}(x) = \sqrt[m-1]{x + \sqrt[m]{x}} > 1$$ It is clear for such $m$, we will have $\phi_{2,m}(x) \ge 1$.

Recall for any $k > 1$ and $t > 0$, $\sqrt[k]{1 + t} < 1 + \frac{t}{k}$.

Start from $\phi_{m,m}(x) = \sqrt[m]{x} \le 1$, we have

$$\begin{align} & \phi_{m-1,m}(x) = \sqrt[m-1]{x + \phi_{m,m}(x)} \le \sqrt[m-1]{x + 1} \le 1 + \frac{x}{m-1}\\ \implies & \phi_{m-2,m}(x) = \sqrt[m-2]{x + \phi_{m-1,m}(x)} \le \sqrt[m-2]{x + 1 + \frac{x}{m-1}} \le 1 + \frac{1}{m-2}\left(1 + \frac{1}{m-1}\right)x\\ \implies & \phi_{m-3,m}(x) = \sqrt[m-3]{x + \phi_{m-2,m}(x)} \le 1 + \frac{1}{m-3}\left(1 + \frac{1}{m-2}\left(1 + \frac{1}{m-1}\right)\right)x\\ & \vdots\\ \implies & \phi_{2,m}(x) \le 1 + \frac12\left( 1 + \frac13\left(1 + \cdots \left(1 + \frac{1}{m-1}\right)\right)\right)x \le 1 + (e-2)x \end{align} $$

Notice for fixed $x$ and as a sequence of $m$, $\phi_{2,m}(x)$ is monotonic increasing. By arguments above, this sequence is ultimately sandwiched between $1$ and $1 + (e-2)x$. As a result, $\phi_{2,\infty}(x)$ is defined for this $x$ and satisfies

$$1 \le \phi_{2,\infty}(x) \le 1 + (e-2) x$$

Taking $x \to 0^{+}$, we get

$$1 \le \liminf_{x\to 0^+} \phi_{2,\infty}(x) \le \limsup_{x\to 0^+}\phi_{2,\infty}(x) \le \limsup_{x\to 0^+}(1 + (e-2)x) = 1$$ This implies $\lim\limits_{x\to 0^+} \phi_{2,\infty}(x)$ exists and equal to $1$.