Working in $\mathbb{Z}_p[x]$, you have the "Freshman's Dream" binomial theorem:
$$(a+b)^p = a^p + b^p.$$
(In fact, this holds in any commutative ring of characteristic $p$, simply by the Binomial Theorem, as mentioned by Qiaochu: all the mixed coefficients are multiples of $p$).
So $(x-c)^p = x^p - c^p$. By Fermat's Little Theorem, you also know that $c^p \equiv c \pmod{p}$, so $(x-c)^p = x^p - c$ in $\mathbb{Z}_p[x]$. That's pretty much all you need.
Addendum
Let me add that you were very perceptive to notice that even though the function $x\mapsto (x-c)^p$ is the same as the function $x\mapsto x-c$ in $\mathbb{Z}_p$, this does not mean that the polynomials $(x-c)^p$ and $x-c$ are equal. We are used to (in the integers, rationals, reals, etc) the fact that two polynomials give the same function if and only if they are identical, but this does not occur over finite rings; for example, Fermat's Little Theorem tells you that the polynomial $x^p$ and the polynomial $x$ both define the same function.
However, what you can prove is that if $p(x)$ and $q(x)$ are two polynomials with coefficients in $\mathbb{Z}_p$, and they define the same function (that is, $p(a) = q(a)$ for all $a\in\mathbb{Z}_p$), then they differ by a multiple of the polynomial $x^p-x$; i.e., $p(x)\equiv q(x)\pmod{x^p-x}$ in $\mathbb{Z}_p[x]$. What you have here is a special case. See if you can figure out how to prove the more general statement.