Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$. [duplicate]

For an odd prime, prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$.

I have proved the other way round that any primitive root of $p^n$ is also a primitive root of $p$ but I have not been able to solve this one. I have tried the usual things that is I have assumed the contrary that there does not exist the primitive root following the above condition and then proceeded but couldn't solve it.
Please help.

Let $g$ be a primitive root $\pmod{p^2}$. Then $p|(g^{p-1}-1)$ by Fermat's little theorem and $p^2 \nmid (g^{p-1}-1)$ since $g$ is a primitive root. Thus by Lifting the Exponent Lemma, $p^{n-1}\|((g^{p-1})^{p^{n-2}}-1)$ and $p^n\|((g^{p-1})^{p^{n-1}}-1)$, so $g$ is also a primitive root $\pmod{p^n}, n>1$.

Edit: More details: Let $d$ be the order of $g \pmod{p^n}, n>1$. Since $g$ is a primitive root $\pmod{p^2}$, we have $p(p-1) \mid d$. By above, $d|p^{n-1}(p-1), d\nmid p^{n-2}(p-1)$, so $d=p^{n-1}(p-1)$, and thus $g$ is a primitive root $\pmod{p^n}, n>1$.

We know from here, if $ord_pa=d, ord_{(p^2)}a= d$ or $pd$

If $a$ is a primitive root $\pmod {p^2}, ord_{(p^2)}a=\phi(p^2)=p(p-1)$

Then $ord_pa$ can be $p-1$ or $p(p-1)$

But as $ord_pa<p, ord_pa$ must be $p-1=\phi(p)\implies a$ is a primitive root $\pmod p$

Again from here, $$ord_{(p^s)}(a)=d, ord_{p^{(s+1)}}(a)=pd,\text{ then } ord_{p^{(s+2)}}(a)=p^2d$$

So, as $ord_pa=p-1,ord_{(p^2)}a=p(p-1); ord_{(p^3)}a$ will be $p\cdot p(p-1)=\phi(p^3)$

Again as, $ord_{(p^2)}a=p(p-1) ord_{(p^3)}a=p\cdot p(p-1);ord_{(p^4)}a$ will be $p\cdot p^2(p-1)=\phi(p^4)$