The identity element of a compact group is a limit point of any "polynomial sequence" Is there an "elementary" (say ultrafilter-free) proof of the following fact: if $G$ is a compact (Hausdorff) topological group, if $g \in G$ is any element from this group, and if $P$ is a polynomial with integer coefficients without constant term, then the identity element of $G$ is a limit point of the sequence $n \mapsto g^{P(n)}$.
An other question: for which integer-valued sequences $u_n$ is the result above still true with $P(n)$ replaced by $u_n$, whatever $G$ and $g$ are?
 A: If $\mathcal{H}$ is a van der Corput set of positive integers, then the closure of $\{ g^h\mid h\in\mathcal{H} \}$ contains the identity element. This generalizes the statement in the original post, because $\{ P(n)\mid n>0 \}$ is a van der Corput set for $P\in\mathbb{Z}[x]$ without a constant term. 
By Terry Tao's remark above, it suffices to show that for any $0<\epsilon\leq 1/2$ and for any $\theta_1,\dots,\theta_n\in\mathbb{R}$ there is $h\in\mathcal{H}$ such that $\|h\theta_1\|,\dots,\|h\theta_n\|<\epsilon$. We follow closely the proof of Theorem 9 in Chapter 2 of Montgomery: Ten lectures on the interface between analytic number theory and harmonic analysis. By the earlier results in the chapter, there is a cosine polynomial
$$ T(x) = a_0 + \sum_{h\in\mathcal{H}} a_h \cos 2\pi hx $$
with real coefficients, nonnegative values, $a_0<\epsilon^n$, and $T(0)=1$. Put 
$$ f(x):=\max(0,1-\|x\|/\epsilon), $$
and consider the expression
$$ a_0 + \sum_{h\in\mathcal{H}} a_h f(h\theta_1)\dots f(h\theta_n). $$
It suffices to show that this expression exceeds $\epsilon^n$, because then $f(h\theta_1)\dots f(h\theta_n)\neq 0$ follows for some $h\in\mathcal{H}$. The Fourier expansion
$$ f(x) = \sum_{k\in\mathbb{Z}}\hat f(k) e(kx) $$
converges absolutely, hence the above expression equals
$$ \sum_{h\in\mathcal{H}\cup\{0\}} a_h 
\left(\sum_{k_1\in\mathbb{Z}}\hat f(k_1) e(hk_1\theta_1)\right)\dots \left(\sum_{k_n\in\mathbb{Z}}\hat f(k_n) e(hk_n\theta_n)\right)=$$ 
$$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)
\sum_{h\in\mathcal{H}\cup\{0\}} a_h e(hk_1\theta_1+\dots+hk_n\theta_n)=$$
$$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)
\sum_{h\in\mathcal{H}\cup\{0\}} a_h \cos 2\pi h(k_1\theta_1+\dots+k_n\theta_n)=$$
$$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)T(k_1\theta_1+\dots+k_n\theta_n).$$
The term corresponding to $k_1=\dots=k_n=0$ contributes $\epsilon^n$, while all the other terms are nonnegative, hence we are done.
A: Take a look at Theorem C and Proposition 1.10 in the article Polynomial Extensions of Van Der Waerden´s and Szemeredi´s Theorems by Bergelson and Leibman. I think they prove a lot more than what you need and still the proofs are completely elementary (although quite long and involved).
A: For getting the every-point statement, at-least in the compact abelian case (see Tao's comment above), one can either prove it by harmonic analytic approach (Weyl's equi. criterion + van der corput trick, just like in GH's proof), or one can use a dynamical approach (either topological dynamics or ergodic theoretic approach) by using induction on the degree of the polynomial and a skew-product theorem à-la Furstenberg.
Notice that an almost-every point statement (wrt the Haar measure) is true in a much more general settings by Bourgain's ergodic theorem. 
[Bourgain's theorem says even more, as an ergodic theorem, for example it says something about repetitions to (nhbds of-) the identity].
