Let $m=p^t$ where p is a prime. Prove that $a^{\phi(m)+t} \equiv a^t \bmod{m}$ for ${\bf all}$ integers a

Question

So, I was thinking that $a^{\phi(m)}\equiv 1 \bmod{m}$, thus when multiplying $a^t$ on both sides, we get that $a^{\phi(m)+t} \equiv a^t \bmod{m}$. What is throwing me off is the all integers a part.

Answer 1

$$a^{\phi(p^t)+t}-a^t=a^t(a^{\phi(p^t)}-1)$$

Now as $p$ is prime; either $p$ divides $a$ or $p\not|a\iff (p,a)=1$

If $p$ divides $a, p^t|a^t$

Else $(p,a)=1$ which can safely use Euler's Totient Theorem