Euler phi function proof

Question

Let $a,e$ and $f \in \Bbb{Z}^+$ and p is prime number. If $\phi(p^e) | f$ then show $a^f \equiv 1 \pmod{p^e}$ $(gcd(a,p^e)=1$

Answer 1

A more generalized version is: given $a,f,n$ such that $(a,n)=1$ and $ \phi(n) |f$ then $a^f = 1 \mod n$

One definition of $ \phi(n) = \# \{ x : (x,n)=1, 1\leq x\leq n\}$ and so the main idea is to prove that $ \{ x : (x,n)=1, 1\leq x\leq n\} = \{ a x \mod n : (x,n)=1, 1\leq x\leq n\}$ or equivalently that $\{ a x \mod n : (x,n)=1, 1\leq x\leq n\}$ is a permutation of $\{ x : (x,n)=1, 1\leq x\leq n\} $

And so $x_1 \cdots x_\phi(n) = a x_1 \cdots a x_\phi(n) = a^{\phi(n)} x_1 \cdots x_\phi(n) \mod n$ and so $ a^{\phi(n)} = 1 \mod n$ since all $x_i$ have inverses mod $n$ which gives you that $ a^f = 1 \mod n$ since $f = e \phi(n) $ and so $ a^f = a^{\phi(n) e } = (a^{\phi(n)})^e = 1^e = 1 \mod n$

This is pseudo-proof.