How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1?

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.