Hey!!!

I have the following exercise:

If $p$ is prime, $p \nmid a$, $p \nmid b$, prove that $$a^p \equiv b^p \pmod p \Rightarrow a^p \equiv b^p \pmod {p^2}$$

My idea is the following:

$$ a^p \equiv b^p \pmod p $$

$$ a^p \equiv a \pmod p $$

$$ b^p \equiv b \pmod p $$

$$ a \equiv b \pmod p \Rightarrow a=b+kp, k \in \mathbb{Z}$$

$$ a^p=(b+kp)^p=\sum_{m=0}^{p} \binom{p}{m} (kp)^mb^{p-m}=\binom{p}{0}b^p+\binom{p}{1}(kp)b^{p-1}+ \dots + \binom{p}{p}(kp)^p= \\ b^p+kp^2b^{p-1}+ \frac{1}{2}(p-1)p^3b^{p-2}k^2+\dots + k^pp^p$$

$$ p^2 \mid kp^2b^{p-1}+ \frac{1}{2}(p-1)p^3b^{p-2}k^2+\dots + k^pp^p

\Rightarrow p^2 \mid a^p-b^p $$

Is this correct??