Prove that $a\equiv b\pmod p\Rightarrow a^p \equiv b^p \pmod {p^2}$ Suppose $p \nmid b$ and $p \nmid a$. Let $a^p \equiv b^p (\text{mod}(p))$. Prove that $a^p \equiv b^p (\text{mod}(p^2))$. 
Below is my proof. I'm interested in two things. 
1) Verifing this proof
2) Finding a simpler proof (preferably one that doesn't involve the Binomial Theorem).
Suppose $a^p \equiv b^p (\text{mod}(p))$. 
Since $a^{p - 1} \equiv b^{p - 1} \equiv 1 (\text{mod}(p))$ By Fermat's Little Theorem, we have $a^p \equiv a \equiv b^p \equiv b (\text{mod}(p))$. 
Since $a \equiv b (\text{mod}(p))$ we can write $ a = pk + n$ and $b = pm + n$ where $k$ and $m$ are non-negative integers and $ 1 \leq n \leq p - 1$. 
We then have $a^p = (pk + n)$ and $b^p = (pm + n)^p$. 
Using the Binomial Theorem we can expand these. Looking at these expanded binomials modulo $p^2$ we see that any terms with powers of $p$ larger than 2 are 0 modulo $p^2$. Thus we are only concerned with the last two terms in the expansion. The expansions are displayed below:  
$a^p = (pk + n)^p \equiv \frac{p!}{(p-1)!}n^{p-1}(pk) + n^p \equiv n^p\mod{p^2}$ 
and 
$b^p = (pm + n)^p \equiv \frac{p!}{(p-1)!}n^{p-1}(pm) + n^p \equiv n^p\mod{p^2}$
Thus $a^p \equiv b^p \mod{p^2}$
I would really appreciate any critique on the proof including the formatting. Thanks.

Simpler proof:
By Fermat $p \ |  \ a-b$, then $a \equiv b$ in modulo $p$
\begin{align*}
\frac{a^p-b^p}{a-b} &= (a^{p-1} + a^{p-2}b + \cdots b^{p-2}a + b^{p-1})\\
&\equiv (b^{p-1} + b^{p-2}b + \cdots b^{p-2}b + b^{p-1})\\
&\equiv p \ b^{b-1} \equiv 0
\end{align*}
in modulo $p$
Since $\frac{a^p-b^p}{a-b}$ and $a-b$ are divisible by $p$. The product $a^p-b^p$ is divisible by $p^2$.
