How prove this polynomial $p(x)$ is deg greater than $n-1$ Question:

Let $P(x)$ be a polynomial satisfying $$P(k)=\cos{\dfrac{2k\pi}{n}},k=1,2,\cdots,n$$
Show that
$$\deg{P(x)}\ge n-1$$

I want to consider
$$P(m+1)-P(m)=\cos{\dfrac{2(m+1)\pi}{n}}-\cos{\dfrac{2m\pi}{n}}=2\sin{\dfrac{\pi}{n}}\sin{\dfrac{2m+1\pi}{n}}$$
Maybe this can use   mean value theorem for divided differences  Or can we also prove
$$P^{(n-1)}(x)\neq 0?$$ But I can't figure out how. Thank you.
 A: Let $\Delta$ be the finite difference operator, $\Delta f(x)=f(x+1)-f(x)$.
Then, as you've shown,
$$\Delta P(m) = -2 \sin \frac{\pi}{n}\sin \frac{(2m+1)\pi}{n}$$
whenever $1\leq m \leq n-1$.
Similarly, we have
$$
\Delta^2P(m)=-2\sin\frac{\pi}{n}\left(\sin\frac{(2m+3)\pi}{n}-\sin\frac{(2m+1)\pi}{n}\right)=-2\sin\frac{\pi}{n}\left(2\sin\frac{\pi}{n}\cos\frac{(2m+2)\pi}{n}\right)
$$
for $1 \leq m \leq n-2$;
$$
\Delta^3P(m)=-4\sin^2\frac{\pi}{n}\left(\cos\frac{(2m+4)\pi}{n}-\cos\frac{(2m+2)\pi}{n}\right)=8\sin^3\frac{\pi}{n}\sin\frac{(2m+3)\pi}{n}
$$
for $1 \leq m \leq n-3$; and so on.
Continuing in this manner, it is convenient to define 
$$
\operatorname{trig}_k m=\begin{cases}
\cos m && k\text{ even}\\
\sin m && k\text{ odd}
\end{cases}$$
Then it follows by induction that
$$
\Delta^k(m)=(-1)^{\lceil k/2 \rceil}2^k\sin^k\frac{\pi}{n}\operatorname{trig}_k \frac{(2m+k)\pi}{n}
$$
for $1 \leq m \leq n-k$.
In particular,
\begin{eqnarray}
\Delta^{n-1}(1)&=&(-1)^{\lceil (n-1)/2 \rceil}2^{n-1}\sin^{n-1}\frac{\pi}{n}\operatorname{trig}_{n-1}\frac{(n+1)\pi}{n}\\
&=&(-1)^{\lceil (n+1)/2 \rceil}2^{n-1}\sin^{n-1}\frac{\pi}{n}\operatorname{trig}_{n-1}\frac{\pi}{n}
\end{eqnarray}
can easily be seen to be nonzero for $n>1$: $\sin \frac{\pi}{n}$ only vanishes when $n=1$, $\cos \frac{\pi}{n}$ only vanishes when $n=2$, and when $n=2$ the formula involves $\operatorname{trig}_1 \frac{\pi}{2}=\sin \frac{\pi}{2}$.
If $n=1$, the problem statement is trivially true. But if $n>1$, we have just shown that the $(n-1)^{\rm st}$ finite difference of $P$ doesn't vanish everywhere. It follows that, if $P$ is a polynomial, it must have degree at least $n-1$.
