I'd like some advice on my approach for this exercise, which has been giving me some hesitancy. Let $\alpha$ be a primitive root modulo $p^n$. Show that $\alpha$ is also a primitive root modulo $p$, where $p$ is an odd prime. (Note that the odd prime distinction is made since $2$ needs to be treated different than the rest of the primes.) Here's what I have.
Since $\alpha$ is a primitive root modulo $p^n$, $\alpha^{\varphi(p^n)}\equiv1\pmod{p^n}$. It is evident $\varphi(p^n)=p^n-p^{n-1}=p^n\cdot(1-1/p)=p^n\cdot\left(\frac{p-1}{p}\right)=\left(\frac{p^n}{p}\right)\cdot(p-1)=p^{n-1}\cdot(p-1)$. Now, $\alpha^{p^{n-1}\cdot(p-1)}\equiv1\pmod{p^n}$. Observe that a primitive root modulo $p$, $x$, implies $x^{\varphi(p)}=x^{p-1}\equiv1\pmod{p}$. Since $p^{n-1}\cdot(p-1)$ is a multiple of $p-1$, we conclude $\alpha$ is also a primitive root modulo $p$. $Q.E.D.$
My only concern is the final step. I don't feel simply claiming $p^{n-1}\cdot(p-1)$ is a multiple of $p-1$ is a strong enough justification between modulo $p^n$ and modulo $p$. Am I just overthinking this proof or is there something I can adjust?
We must show that the powers of $\alpha$ cover all residue classes $\bmod p$ that are coprime to $p$ ( this is equivalent to showing $\alpha$ generates the multiplicative group $\bmod p$, or that $\alpha$ is a primtive root $\bmod p$).
So take $a$ coprime to $p$, since $\alpha$ is a primitive root $\bmod p^n$ and $\gcd(a,p^n)=1$ there is $k$ such that $\alpha^k \equiv a \bmod p^n$, it follows $\alpha^k \equiv a \bmod p$ as desired.