alternative-prooffield-theoryfinite-fields
If $p$ is prime, then $(x+y)^p=x^p+y^p$ holds in any field of characteristic $p$. However all the proofs I have seen use induction and some relatively nasty algebra despite how fundamental this fact seems.
What is the nicest, "highest level proof" you know?
Here are three phenomena specific to the positive characteristic case which I have found to lead to many of the differences from characteristic zero.
1: Let $k$ be a field, let $k[t]$ be the univariate polynomial ring over that field, and consider the derivative $\frac{d}{dt}$, i.e., the unique $k$-derivation on $k[t]$ which maps $t$ to $1$. The derivative is in particular a $k$-linear map, so it has a kernel. In characteristic zero the kernel is just the subring $k$ of constant polynomials, a very small and well-understood subspace. But in characteristic $p > 0$ the kernel consists of $k[t^p]$, i.e., all polynomials of the form $f(t^p)$, which is a big, infinite dimensional subring abstractly isomorphic to the original ring. Therefore having zero differential means something subtly, but profoundly, different in positive characteristic. Both separability of field extensions and (more generally) etaleness of maps can be characterized in terms of differentials, so this leads to a lot of the well-known differences. But it comes up in other places too: for instance, the theory of Weierstrass points on algebraic curves in positive characteristic is totally different and apparently less satisfactory: for a curve in characteristic zero of genus $g \geq 2$, the Weierstrass points form a set of $g^3-g$ points. However in characteristic $p$ it is possible for every point to be a Weierstrass point! The problem it seems is that early on in the theory of Weierstrass points one needs to write down a Wronskian, i.e., some matrix of derivatives. This can go haywire in positive characteristic.
1bis: Similarly in positive characteristic there are nonlinear polynomials whose derivative is everywhere nonzero: e.g. the Artin-Schreier map $x^p - x$. Thus in characteristic $p$ the affine line is not simply connected. In general there are many more unramified coverings of affine varieties in characteristic $p$ than in characteristic zero.
2: In any ring of characteristic $p > 0$, the map $x \mapsto x^p$ is an endomorphism, the Frobenius map. This gives everything in positive characteristic an extra structure that is not present in characteristic zero: for instance it gives rise to "new" isogenies of abelian varieties, and so forth. Here is a very low brow exploitation of the Frobenius map on $R = \mathbb{Z}/p\mathbb{Z}$: clearly $1^p = 1$, and by the homomorphic property of Frobenius, $2^p = (1+1)^p = 1^p + 1^p = 2$. In general, if $x \in R$ is such that $x^p = x$, then $(x+1)^p = x^p + 1^p = x+1$. Thus $x^p = x$ for all $x$ in $R$: we have proved Fermat's Little theorem by induction on $x$!
3: There exist finite fields of characteristic $p$, whereas any field of characteristic $0$ must contain $\mathbb{Q}$ and thus be infinite. Working in a finite field can be strange, because any given nonzero univariate polynomial over a field vanishes at only finitely many points. When the field is infinite, this is a very small set, but over a finite field there are nonzero polynomials which vanish at every point. Another way to say this is that the map which associates to $f \in k[t]$ its function $x \mapsto f(x)$ is injective iff $k$ is infinite (it is also surjective iff $k$ is finite!). When $k$ is finite of cardinality $p$, the kernel is the ideal generated by $t^p-t$. There's that Artin-Schreier map again! These considerations actually lead swiftly to a proof of the Chevalley-Warning Theorem.
3bis: a more geometric way of expressing the above is that a field $k$ is infinite iff the set of $k$-rational points of affine $n$-space over $k$ is Zariski dense in the whole space. When this does not hold (i.e., when $k$ is finite), "general position" arguments can fail dramatically, so that often in big theorems finite fields need to be treated separately from infinite fields. Sometimes the simple structure of finite fields leads to alternate, easier proofs -- e.g. in the Primitive Element Theorem or the existence of separable splitting fields for central simple algebras -- but sometimes not: e.g. the proof of Noether Normalization is much nastier over a finite field, to the extent that some texts retreat to the case of an algebraically closed field (when what the standard proof needs is precisely that the field be infinite). Finally, Bjorn Poonen rather recently came up with a useful reparation for Bertini type theorems over finite fields: one cannot intersect with a hyperplane because there are not "enough" $k$-rational hyperplanes. Rather one needs to intersect with a hypersurface of relatively low degree.
Consider a field $F$ of characteristic $p$. A polynomial has multiple roots only if it has a root in common with its (formal) derivative; that is, the multiple roots of $f$ are the roots of $\gcd(f,f')$. Since $f$ is irreducible, multiple roots can occur only if the $\gcd$ is $f$ itself, that is $f'$ is a multiple of $f$. And that is only possible if $f'=0$, that is, all monomials in $f$ have degree a multiple of $p$, so $f(x)=g(x^p)$ for some polynomial $g$.
If $F$ is finite, then $\phi\colon a\mapsto a^p$ is an automorphism of $F$ (and also of the splitting field $E$ of our polynomial), and there exists $h(x)$ such that $\phi(h)=g$. Then for $\alpha\in E$ with $f(\alpha)=0$ also $h(\alpha)=0$ (because $\phi(h(\alpha))=\phi(h)(\phi(\alpha))=g(\alpha^p)=f(\alpha)=0$). Since $h$ is of smaller degree than $f$, $f$ is not irreducible.
As this proof shows, one has to look for cases where $\phi$ is not an automorphism to find a counterexample (such as in Andreas Carantis comment).
Best Answer
The binomial coefficient $\binom p i$ is divisible by $p$ for $1 \leq i \leq p-1$
One way of seeing this is Legendre's formula on the power of a prime dividing some factorial, http://www.cut-the-knot.org/blue/LegendresTheorem.shtml
and http://en.wikipedia.org/wiki/Factorial#Number_theory
From the formula, $p$ divides $p!$ with exponent exactly $1,$ but $p$ does not divide $i!$ or $(p-i)!$ when $1 \leq i \leq p-1.$