Binomial formula modulo a prime. Let $p$ be a prime. I see that modulo $p$, the binomial formula reduces to 
$$(x+y)^p\equiv x^p+y^p  \pmod p$$ because $p \choose k$ is a multiple of $k$ whenever $k=1..p-1$,  but don't we have by Fermat's little theorem 
that $a^p\equiv a \pmod p$ for any integer $a$ and so this is true for $a=x+y$
hence the binomial formula reduces to 
$(x+y)^p\equiv x+y \pmod p$. Thanks for your help!
 A: Adding on to my comment:
Yes, everything that you have written is true. It is the case that
$$(x+y)^p \equiv x^p+y^p \mod p$$
using the Binomial Theorem, and it is also true that
$$(x+y)^p \equiv x+y \mod p$$
by Fermat's Little Theorem.
There is not contradiction here because by Fermat's Little Theorem, we have that
$$x^p \equiv x \mod p \quad\text{ and }\quad y^p \equiv y \mod p$$
and so
$$x^p+y^p\equiv x+y \mod p$$
as we expect.
Your observation that
$$(x+y)^p \equiv x^p+y^p \mod p$$
by the Binomial Theorem is sometimes used to prove Fermat's Little Theorem by induction. The (complete) proof is as follows:
First we show, as noted by you, that $p \mid \binom{p}{k}$ for $1 \leq k < p$. We can prove this either by considering the factors of $p$ in the numerator and denominator of $\binom{p}{k}$, or by noticing that
$$ \binom{p}{k} = \frac{p}{k}\binom{p-1}{k-1}$$
Thus
$$ k \mid p\binom{p-1}{k-1}$$
and since $k$ and $p$ are relatively prime, we get that
$$ k \mid \binom{p-1}{k-1}$$
showing that
$$ \frac{1}{k}\binom{p-1}{k-1}$$
is an integer, and so
$$ \binom{p}{k} = p \cdot \frac{1}{k}\binom{p-1}{k-1}$$
is an integer divisible by $p$.
As you noted in your question,
$$(x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k} = x^p + y^p + \sum_{k=1}^{p-1} \binom{p}{k} x^k y^{p-k} $$
Since every term in the final sum is divisble by $p$, we get that
$$(x+y)^p \equiv x^p + y^p \mod p$$
and this holds for all integers $x$ and $y$.
We can now prove by induction that
$$x^p \equiv x \mod p$$
for all integers $x$.
The base case of $x=0$ is straightforward since $0^p=0$.
Now suppose that $x^p \equiv x \mod p$ for some $x$. Then we have that
$$ (x+1)^p \equiv x^p + 1^p \equiv x+1$$
using our observation above, and the inductive hypothesis.
Thus the claim holds for $x+1$ as well, and hence for all natural numbers by induction. For the negative integers, we can note that if $x$ is a positive integer then
$$ (-x)^p = (-1)^p x^p \equiv -x \mod p$$
Here, we use that $x^p \equiv x \mod p$, and also that $(-1)^p \equiv -1 \mod p$, since if $p$ is odd then $(-1)^p = -1$, and if $p=2$ then $(-1)^p = 1 \equiv -1 \mod 2$.
A: Yes, good point. If $x$ and $y$ are integer, we can simply write $(x+y)^p = x + y \mod p$.
However, the formula $(x+y)^p \equiv x^p + y^p \mod p$ is also interesting because it applies not only to elements of $\mathbb{Z}/p\mathbb{Z}$, but also to any commutative algebra over $\mathbb{Z}/p\mathbb{Z}$ (for example, you may apply it to polynomials with coefficients in $\mathbb{Z}/p\mathbb{Z}$, or for field extensions of $\mathbb{Z}/p\mathbb{Z}$).
A: You have to show that $p \mid \binom{p}{k}$ when $1 < k < p$. We have: $k!(p-k)! \mid p!=p(p-1)!$ Using unique prime factorization we can write:
$k!(p-k)! = a_1^{n_1}\cdot a_2^{n_2}\cdots a_m^{n_m}$ with the $a_i$'s are distinct primes smaller than $p$, and none of the factors divide $p$ since $a_i\nmid p$. Thus $a_i^{n_i} \mid (p-1)!$,and since these factors are pairwise coprime, we have $k!(p-k)! \mid (p-1)!$, this shows $p \mid \binom{p}{k}$, and the conclusion follows from this fact.
