Binary, decimal and hexadecimal converting

The answer is when we converting from binary to hexadecimal, $(\overline{m_1m_2m_3\cdots m_{4n}})_2\rightarrow (\overline{k_1k_2k_3\cdots k_n})_{16}$ , where $m_n$ is only one digit, then we can convert 4 digits at a time from the end of the number, because $2^4=16$ . So we just do the same thing back and forth, when we convert the number to binary and this number to hexadecimal.

Note: The max value of $\overline{m_{4n-3}m_{4n-2}m_{4n-1}m_{4n}}$ is $2^3+2^2+2^1+2^0=15$ , so $\overline{m_{4n-3}m_{4n-2}m_{4n-1}m_{4n}}$ can influence only the last digit of the hexadecimal number.

I know I can prove it , but I don't know how. :) (Also the earth isn't flat, its an oblate spheroid ). XD

I can't prove that, but the earth isn't flat. :)