\(\displaystyle{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\leq{\left({\frac{{{1}+{1}}}{{{2}}}}\right)}^{{n}}={1}\)

Now \(\displaystyle{\cos{{u}}}\geq{1}-{\frac{{{u}^{{2}}}}{{{2}}}}\) for small \(\displaystyle{u}\geq{0}\), then

\(\displaystyle{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\geq{\left({1}-{\frac{{{1}}}{{{2}^{{{2}{\left({n}+{1}\right)}}}}}}\right)}^{{n}}\to{1}\)

Now \(\displaystyle{\cos{{u}}}\geq{1}-{\frac{{{u}^{{2}}}}{{{2}}}}\) for small \(\displaystyle{u}\geq{0}\), then

\(\displaystyle{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\geq{\left({1}-{\frac{{{1}}}{{{2}^{{{2}{\left({n}+{1}\right)}}}}}}\right)}^{{n}}\to{1}\)