Example 6 Consider the double-angle cosine formula $\cos 2 \alpha=2 \cos ^{2} \alpha-1$,
First, transform to get: $2 \cos 2 \alpha+1=4 \cos ^{2} \alpha-1=(2 \cos \alpha+1)(2 \cos \alpha-1)$, and write it as a recursive split term: $2 \cos 2^{k} \alpha-1=\frac{2 \cos 2^{k+1} \alpha+1}{2 \cos 2^{k} \alpha+1}, k \in \mathbf{N}$. This yields the equation: $(2 \cos \alpha-1)(2 \cos 2 \alpha-1)(2 \cos 4 \alpha-1) \cdots\left(2 \cos 2^{n-1} \alpha-1\right)=\frac{2 \cos 2^{n} \alpha+1}{2 \cos \alpha+1}$.