How I show that $(b-a)^p \equiv b^p-a^p \mod{p}$? 
If $b \ge a$ and p is prime, then  $(b-a)^p \equiv b^p-a^p \mod{p}$.

I know this: 
$b^p = (b - a+a)^p \equiv (b-a)^p +a^p \mod{p} $. 
After all help in this comments I wrote this, I want to know if is correct my answer like this?
\begin{align}
b^p = (b -a +a)^p \equiv (b-a)^p +a^p \mod{p}
\end{align}
\begin{align}
b^p \equiv (b-a)^p +a^p \mod{p}
\end{align}
So, we have that:
\begin{align} 
b^p\equiv a^p \mod{p} \leftrightarrow p | b^p -a^p
\end{align}
and, 
\begin{align} 
b-a^p \equiv b -a \equiv b^p -a^p\mod{p} \to\\
(b-a)^p\equiv b^p-a^p \mod{p} \to \\
p| (b-a)^p - (b^p-a^p) \to \\
\end{align}
$p|(b-a)^p \to p|(b-a)$ and $p|(b^p-a^p)$
So, we have that 
\begin{align}
p | (b-a)^p - (b^p -a^p) \to (b-a)^p \equiv b^p -a^p.
\end{align}


I know this: $b^p = (b - a+a)^p \equiv (b-a)^p +a^p \mod{p}$

Well if you know that then just subtract $a^p$ from both sides:
$b^p -a^p\equiv (b-a)^p +a^p -a^p \equiv (b-a)^p\mod{p}$
......
But perhaps more simply:
By binomial expansion $(n + m)^k \equiv n^k + m^k \mod c$ so 
$(b -a)^p \equiv b^p + (-a)^p\equiv b^p\pm a^p\mod p$.  (With $\pm$ being $+$ if $p$ is even and $\pm$ being $-$ if $p$ is odd.  And if $p = 2$ then $\pm$ could be either as $1 \equiv -1\mod 2$.)
As $p$ is prime, either $p$ is odd or $p$ is $2$. 
