TITLE: Prove that $\inf\limits_{n\in\mathbb N}\sum\limits_{k=0}^{p}\lvert\sin{(n+k)^p}\rvert>0$
QUESTION [12 upvotes]: For any positive integer number $p$, show that
$$\inf\left\{
{\left\vert\sin{(n^p)}\right\vert+\left\vert\sin{(n+1)^p}\right\vert+\cdots+
\left\vert\,\sin{(n+p)^p}\right\vert\, :\,n\in \mathbb{N}}\right\}>0.
$$
My try. I only can prove for $p=2$. I have proved the following
$$\vert\sin{(n+1)^2}\vert+\vert\sin{(n-1)^2}\vert+\vert\sin{n^2}\vert\ge\dfrac{1}{2}\sin{2}$$
Lemma. for any  $x,y$ we have
$$
\lvert\sin{x}\rvert+\lvert\sin{y}\rvert\ge\lvert\sin{(x-y)}\rvert.
$$
This lemma is easy to prove.
Proof:
\begin{align*}
&|\sin{(n+1)^2}|+|\sin{(n-1)^2}|+|\sin{n^2}|\ge\dfrac{1}{2}[\sin{(n+1)^2}|+|\sin{(n-1)^2}|+2|\sin{n^2}|]\\
&\ge\dfrac{1}{2}(|\sin{[(n+1)^2-n^2]}|+|\sin{[n^2-(n-1)^2]}|)\\
&\ge\dfrac{1}{2}|\sin{[(n+1)^2+(n-1)^2-2n^2]}|=\dfrac{1}{2}\sin{2}
\end{align*}
But my problem I can't it,
This problem is from a analysis problem book. This author's only hint, we note this 
$$\sum_{i=0}^{p}(-1)^i\binom{p}{i}(n+p-i)^p=p!$$
But I can't. Thank you

REPLY [4 votes]: An another proof if someone is interested,

$(X^p,..., (X+p)^p)$  is a base of ${\mathbb{Q}}_p [X]$: There exist $m\in \mathbb{N^*}$ and $\lambda_0 , ... , \lambda_p \in \mathbb{Z^{p+1}}$ such that $m = \lambda_0 X^p + \lambda_1 (X+1)^p + ... + \lambda_p (X+p)^p$.
Let's assume that 
$$
\inf_{n\in\mathbb N}|\sin{(n^p)}|+|\sin{(n+1)^p}|+\cdots+|\sin{(n+p)^p}|=0.
$$
Then there exist a sequence $(n_k)$ such that $\forall j\in [[0,p]]$, 
$$\lim_{k\to +\infty}\sin ((n_k+j)^p )=0$$  i.e.
$$\lim_{k\to +\infty} e^{2i(n_k+j)^p}=1$$ 
Thus, $e^{2im}$ tends to 1

Therefore $m=0$. This is a contradiction, and hence
$$
\inf_{n\in\mathbb N}|\sin{(n^p)}|+|\sin{(n+1)^p}|+\cdots+|\sin{(n+p)^p}|>0.
$$ QED