[Proof] It is not difficult to prove by mathematical induction that when $|x|>2$, there must be $p_{n}(x)>x$. Therefore, we discuss $p_{n}(x)$ on the domain $[-2,2]$, and make the substitution
$$x=2 \cos t, t \in[0, \pi]$$

First, we use mathematical induction to prove that for any natural number $n$,
$$p_{n}(2 \cos t)=2 \cos 2^{n} t$$

In fact, when $n=1$, we have
$$p_{1}(2 \cos t)=2\left(2 \cos ^{2} t-1\right)=2 \cos 2 t$$

which shows that the proposition is correct.
Assume the proposition is correct for $n-1$, i.e.,
$$p_{n-1}(2 \cos t)=2 \cos 2^{n-1} t$$

By the definition of $p_{n}(x)$, we have
$$\begin{aligned} p_{n}(2 \cos t) & =p_{1}\left[p_{n-1}(2 \cos t)\right] \\ & =p_{1}\left(2 \cos 2^{n-1} t\right) \\ & =2\left(2 \cos ^{2} 2^{n-1} t-1\right)=2 \cos 2^{n} t \end{aligned}$$

Thus, the proposition is also correct for $n$.
Next, we solve the equation $p_{n}(2 \cos t)=2 \cos t$,
i.e., solve the equation $2 \cos 2^{n} t=2 \cos t$,
Solving it, we get $2^{n} t= \pm t+2 m \pi$ (where $m$ is an integer).
From the above solutions, we take the following $2^{n}$ values:
$$\begin{array}{l} t_{k}=\frac{2 k \pi}{2^{n}-1}, k=0,1, \cdots, 2^{n-1}-1 \\ s_{l}=\frac{2 l \pi}{2^{n}+1}, l=1,2, \cdots, 2^{n-1} \end{array}$$

Obviously, these two sets of values satisfy the following relationships:
$$\begin{array}{l} 0 \leqslant t_{0}<t_{1}<\cdots<t_{2^{n-1}-1}<\pi \\ 0<s_{1}<s_{2}<\cdots<s_{2^{n-1}}<\pi \end{array}$$

Since the cosine function is monotonically decreasing in the interval $[0, \pi]$, it follows that $2 \cos t_{k}$ (correspondingly, $2 \cos s_{l}$),
$k=0,1, \cdots, 2^{n-1}-1$ (correspondingly, $l=1,2, \cdots, 2^{n-1}$) are different solutions of the equation.

Next, we prove that $t_{k 0} \neq s_{l 0}\left(0 \leqslant k_{0} \leqslant 2^{n-1}-1,1 \leqslant l_{0} \leqslant 2^{n-1}\right)$.
If not, from $\frac{2 k_{0} \pi}{2^{n}-1}=\frac{2 l_{0} \pi}{2^{n}+1}$
we get $l_{0}=k_{0} \frac{2^{n}+1}{2^{n}-1}=k_{0}+k_{0} \frac{2}{2^{n}-1}$.
Since $\left(2,2^{n}-1\right)=1$, i.e., 2 and $2^{n}-1$ are coprime, and $l_{0}$ is a natural number, it follows that $2^{n}-1$ must divide $k_{0}$, but $0 \leqslant k_{0} \leqslant 2^{n-1}-1$, so it must be that $k_{0}=0$, thus $l_{0}=0$, which contradicts $1 \leqslant l_{0}$.

Therefore, $2 \cos t_{k 0} \neq 2 \cos S_{l 0}\left(0 \leqslant k_{0} \leqslant 2^{n-1}-1,1 \leqslant l_{0} \leqslant 2^{n-1}\right)$, which means
$$x_{k}=2 \cos t_{k} \quad\left(k=0,1, \cdots, 2^{n-1}-1\right)$$

and $x_{l}=2 \cos S_{l} \quad\left(l=1,2, \cdots, 2^{n-1}\right)$ are $2^{n}$ distinct real solutions of the equation $p_{n}(x)=x$. And by the definition of $p_{n}(x)$, this equation is a $2^{n}$-degree algebraic equation. By the fundamental theorem of algebra, the equation has no other solutions.