How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1? The Weyl equidistribution theorem states that the sequence of fractional parts ${n \xi}$, $n = 0, 1, 2, \dots$ is uniformly distributed for $\xi$ irrational. 
This can be proved using a bit of ergodic theory, specifically the fact that an irrational rotation is uniquely ergodic with respect to Lebesgue measure. It can also be proved by simply playing with trigonometric polynomials (i.e., polynomials in $e^{2\pi i k x}$ for $k$ an integer) and using the fact they are dense in the space of all continuous functions with period 1.  In particular, one shows that if $f(x)$ is a continuous function with period 1, then for any $t$,  $\int_0^1 f(x) dx = \lim \frac{1}{N} \sum_{i=0}^{N-1} f(t+i \xi)$. One shows this by checking this (directly) for trigonometric polynomials via the geometric series.  This is a very elementary and nice proof.
The general form of Weyl's theorem states that if $p$ is a monic integer-valued polynomial, then the sequence ${p(n \xi)}$ for $\xi$ irrational is uniformly distributed modulo 1.  I believe this can be proved using extensions of these ergodic theory techniques -- it's an exercise in Katok and Hasselblatt.  I'd like to see an elementary proof.
Can the general form of Weyl's theorem be proved using the same elementary techniques as in the basic version?
 A: There is a fairly good exposition in Terry Tao's post, see Corollaries 4-6. Here is a sketch:
We prove the more general statement: Let $p(n)= \chi n^d + a_{d-1} n^{d-1} + \cdots + a_1 n + a_0$ be any polynomial, with $\chi$ irrational. Then $p(n) \mod 1$ is equidistributed. Our proof is by induction on $d$; the base case $d=1$ is standard.
Set $e(x) = e^{2 \pi i x}$. By the standard trickery with exponential polynomials, it is enough to show 
$$\sum_{n=0}^{N-1} e(p(n)) = o(N).$$
Choose a positive integer $h$. With a small error, we can replace the sum by
$$\sum_{n=0}^{N-1} (1/h) \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right).$$
By Cauchy-Schwarz, this is bounded by
$$\frac{\sqrt{N}}{h} \left[ \sum_{n=0}^{N-1}  \left( e(p(n))  + \cdots + e(p(n+h-1)) \right) \overline{ \left( e(p(n))  + \cdots + e(p(n+h-1)) \right)} \right]^{1/2}.$$
Expanding the inner sum, we get $h^2$ terms of the form $e(p(n) - p(n+k))$. There are $h$ terms where $k=0$; these each sum up to $N$. For the other $h^2-h$ terms, the sum is of the form $\sum_{n=0}^{N-1} e(q(n))$, where $q$ has leading term $\chi d n^{d-1}$. By induction, each of these sums is $o(N)$. 
So the quantity in the square root is
$$hN+o(N)$$
where the constant in the $o$ depends on $h$ and $\chi$. Putting it all together, we get a bound of
$$N/\sqrt{h} + o(N).$$
Since $h$ was arbitrary, this proves the result.
