Chứng minh rằng 2+2^2+2^3+...+2^89+2^90 chia hết cho 7

$\begin{array}{l} 2 + {2^2} + {2^3} + ... + {2^{89}} + {2^{90}}\\ Co\,\,90\,so\,\,hang\\ A = \left( {2 + {2^2} + {2^3}} \right) + \left( {{2^4} + {2^5} + {2^6}} \right) + ... + \left( {{2^{88}} + {2^{89}} + {2^{90}}} \right)\\ A = 2\left( {1 + 2 + {2^2}} \right) + {2^4}\left( {1 + 2 + {2^2}} \right) + {2^{88}}\left( {1 + 2 + {2^2}} \right)\\ A = 2.7 + {2^4}.7 + .... + {2^{88}}.7\\ \Rightarrow A\,\, \vdots \,\,7 \end{array}$