Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$ 
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$

It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.

Multiply by conjugate:
$$\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}=\\
\lim_{x \rightarrow 0} \frac{1-\cos^2 x\cos^2 2x\cdots \cos^2 nx}{x^2\cdot(1+\cos x\cos 2x\cdots \cos nx)}=\\
\frac12 \lim_{x \rightarrow 0} \frac{1-(1-\sin^2 x)(1- \sin^2 2x)\cdots (1-\sin^2 nx)}{x^2}=\\
\frac12 \lim_{x \rightarrow 0} \frac{(\sin^2 x+\sin^2 2x+\cdots +\sin^2 nx)-(\sin^2 x \sin^2 2x +\cdots)+\cdots}{x^2}=\\
\frac12 \left(1+2^2+\cdots +n^2\right)=\frac{n(n+1)(2n+1)}{12}.$$
Note: The expansion of the brackets is performed by the elementary symmetric polynomial rule (see also here) and the limits of the degrees $4$ and higher when divided by $x^2$ will vanish.
