To prove a fundamental idea is "translation".
First, prove: there exists a primitive root $g$ modulo $p$ such that
$$g^{\rho-1} \not \equiv 1\left(\bmod p^{2}\right)$$

In fact, take any primitive root $g$ modulo $p$. If $g$ does not satisfy (1), we say $g+p$ is a primitive root modulo $p$ that satisfies (1).

First, using the binomial theorem and the fact that $g$ is a primitive root modulo $p$, we know that $g+p$ is a primitive root modulo $p$.

Second, we have
$$\begin{aligned} (g+p)^{\rho-1} & \equiv g^{\rho-1}+p(p-1) g^{\rho-2} \\ & \equiv 1-p g^{\rho-2} \not \equiv 1\left(\bmod p^{2}\right) \end{aligned}$$

Therefore, there exists a primitive root modulo $p$ that satisfies (1).
Next, prove: if $g$ is a primitive root modulo $p$ that satisfies (1), then for any $\alpha \in \mathbf{N}^{*}$, $g$ is a primitive root modulo $p^{\alpha}$.

First, we have the following conclusion: for any $\beta \in \mathbf{N}^{*}$, we can set
$$g^{\alpha\left(p^{\beta}\right)}=1+p^{\beta} \cdot k_{\beta}$$

where $p \nmid k_{\beta}$. In fact, when $\beta=1$, by the choice of $g$, (2) holds. Now assume (2) holds for $\beta$, then
$$\begin{aligned} g^{\rho\left(p^{\beta+1}\right)} & =\left(g^{\rho\left(p^{\beta}\right)}\right)^{p}=\left(1+p^{\beta} \cdot k_{\beta}\right)^{p} \\ & \equiv 1+p^{\beta+1} \cdot k_{\beta}\left(\bmod p^{\beta+2}\right) \end{aligned}$$

Combining $p \nmid k_{\beta}$, we know that (2) holds for $\beta+1$. Therefore, (2) holds for $\beta \in \mathbf{N}^{*}$.
Second, let $\delta=\delta_{p^{\alpha}}(g)$, then by Euler's theorem, we know $\delta \mid p^{\alpha-1}(p-1)$. And by $g$ being a primitive root modulo $p$, and $g^{\delta} \equiv 1\left(\bmod p^{\alpha}\right)$ (of course, also $g^{\delta} \equiv 1(\bmod p)$), we know $(p-1) \mid \delta$. Therefore, we can set $\delta=p^{\beta-1}(p-1)$, where $1 \leqslant \beta \leqslant \alpha$.

Now using (2), we know
$$g^{\rho\left(p^{\beta}\right)} \neq \equiv 1\left(\bmod p^{\beta+1}\right) \text{, i.e., } g^{\delta} \neq 1\left(\bmod p^{\beta+1}\right) \text{.}$$

Combining $g^{\delta} \equiv 1\left(\bmod p^{\alpha}\right)$, we know $\beta \geqslant \alpha$.
In summary, $\beta=\alpha$, i.e.,
$$\delta_{p^{\alpha}}(g)=p^{\alpha-1}(p-1)=\varphi\left(p^{\alpha}\right)$$

Thus, $g$ is a primitive root modulo $p^{\alpha}$.