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

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  • If $g$ is a primitive root of $p^2$ where $p$ is an odd prime, why is $g$ a primitive root of $p^k$ for any $k \geq 1$?

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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 $p\cdot p^2(p-1)=\phi(p^4)$

Let us give a more elementary answer, while still using some binomial theorem. But we shall employ no more than a binomial lemma. It states that, for a prime $p$, and an integer $1\leq a\leq p-1$, we have $p\mid \binom{p}{a}$.
Now you know that $a$ is a primitive root of $p^2$, so the order of $a$ modulo $p^2$ is $p(p-1)$. And we know that $a^{p-1}=1+kp$ for some $k$. Then the assumption that $a$ is a primitive root of $p^2$ implies that $k$ is not divisible by $p$. Therefore $$a^{p^n(p-1)}=(a^{p-1})^{p^n}=1+kp^{n+1}+mp^{n+2}$$ for some $m$.(This is the place where we make use of the binomial lemma.) This directly tells you that $a$ is a primitive root of $p^n$ for every $n\ge1$.

More Details. We elaborate upon two things: Firstly, we show the centered equation, and, secondly, we show how that implies the primitivity of $a$. For the first, use the lemma to deduce that $(a^{p-1})^{p}=1+kp\times p+\text{terms of higher powers of $p$}$. Now, by induction, the result follows. For the second, just divide $(a^{p-1})^{p^r}$ by $p^n$ for $k=0,1,\ldots,n-1$, to see that, for lower powers of $a$ than $(a^{p-1})^{p^{n-1}}$, it cannot be congruent to $1$ modulo $p^n$. So $a$ is indeed a primitive root of $p^n$.

If there is any error in the above proof, please inform me; if there is any ambiguity, please point it out, thanks.