I try to give a proof of the following proposition, but not quite sure if the proof is correct. Can anyone please help verify my proof? Thanks in advance!
If $p$ is an odd prime, $p \not\mid a$, and $p \not\mid n$, then if $x^n \equiv a \pmod p$ is solvable, so is $x^n \equiv a \pmod {p^e}$ for all $e \geq 1$. All these congruences have the same number of solutions.
My proof: Let $d = (n,\phi(p)) = (n,p-1)$. There exists $g\in \mathbb{Z}$ which is simutaneously a primitive root $\mod p$ and a primitive root $\mod {p^e}$. The equation $x^n \equiv a \pmod p$ is solvable iff $a = g^b$ for some integer $b$ with $d \mid b$, and the number of inequivalent solutions is precisely $d$. Moreover, the equation $x^n \equiv a \pmod {p^e}$ is solvable iff $a = g^{b'}$ for some integer $b'$ with $d' = (n,\phi(p^e)) \mid b$, and the number of inequivalent solutions is $d'$. Straightforward calculations show that $d = d'$, which concludes the proof.
edit: Suppose $a=g^b$, $x=g^y$. Then the equation $x^n\equiv a\pmod p$ is equivalent to $g^{ny}\equiv g^{b}\pmod p$, which in turn is equivalent to $ny\equiv b\pmod {\phi(p)}$. The latter is solvable iff $(n,\phi(p))\mid b$. The proof is from Kenneth Ireland's number theory book.