If \(p \nmid a\) and \(p \nmid b\), then:
$$\begin{gather} a^{p-1} ≡ 1 \bmod p \implies a^p ≡ a \bmod p \\ b^{p-1} ≡ 1 \bmod p \implies b^p ≡ b \bmod p\end{gather}$$
If \(a^p ≡ b^p \bmod p\), then \(a ≡ b \bmod p\):
$$ a = b + mp $$
If we raise both sides to the power of \(p\):
$$ a^p = (b + mp)^p $$
Using the binomial theorem:
$$ a^p = {p \choose 0} b^p (mp)^0 + {p \choose 1} b^{p-1} (mp)^1 + {p \choose 2} b^{p-2} (mp)^2 + \cdots + {p \choose {p-1}} b^1 (mp)^{p-1} + {p \choose p} b^0 (mp)^p $$
Since \({p \choose 1} = p\):
$$ a^p = {p \choose 0} b^p (mp)^0 + b^{p-1} m(p^2) + {p \choose 2} b^{p-2} (mp)^2 + \cdots + {p \choose {p-1}} b^1 (mp)^{p-1} + {p \choose p} b^0 (mp)^p $$
Every term other than the first one has a factor of \(p^2\):
$$\begin{align} a^p &= {p \choose 0} b^p (mp)^0 + kp^2 \\ a^p &= b^p + kp^2 \end{align}$$
This means \(a^p ≡ b^p \bmod p^2\).