已知数列 \(\left\{ {{a_n}} \right\}\) ， \({a_n} = {2^n}\) ， \({b_n} = {2^{{a_n}}}\) ， \(M{\rm{ = }}\left( {1 + \frac{1}{{{b_1}}}} \right)\left( {1 + \frac{1}{{{b_2}}}} \right)\left( {1 + \frac{1}{{{b_3}}}} \right) \cdots \left( {1 + \frac{1}{{{b_n}}}} \right)\) ，\(n∈{N}^{*}\)，则（　　）

因 \({a_n} = {2^n}\) ， \({b_n} = {2^{{a_n}}} = {2^{{2^n}}}\) ， \(\left( {1 + \frac{1}{{{b_1}}}} \right)\left( {1 + \frac{1}{{{b_2}}}} \right)\left( {1 + \frac{1}{{{b_3}}}} \right) \cdots \left( {1 + \frac{1}{{{b_n}}}} \right)\) 

\( = \left( {1 + \frac{1}{{{2^2}}}} \right)\left( {1 + \frac{1}{{{2^{{2^2}}}}}} \right)\left( {1 + \frac{1}{{{2^{{2^3}}}}}} \right) \cdots \left( {1 + \frac{1}{{{2^{{2^n}}}}}} \right)\) \( = \frac{{\left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 + \frac{1}{{{2^2}}}} \right)\left( {1 + \frac{1}{{{2^{{2^2}}}}}} \right)\left( {1 + \frac{1}{{{2^{{2^3}}}}}} \right) \cdots \left( {1 + \frac{1}{{{2^{{2^n}}}}}} \right)}}{{1 - \frac{1}{{{2^2}}}}}\) 

\( = \frac{{\left( {1 - {{\left( {\frac{1}{{{2^{{2^n}}}}}} \right)}^2}} \right)}}{{1 - \frac{1}{{{2^2}}}}} < \frac{4}{3} < 2\) .

故选：C.

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