TITLE: Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$.
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: 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.