Prove that for every natural number \(n > 1\) the equality: \[\lfloor n^{1 / 2}\rfloor + \lfloor n^{1/ 3}\rfloor + \dots + \lfloor n^{1 / n}\rfloor = \lfloor \log_{2}n\rfloor + \lfloor \log_{3}n\rfloor + \dots + \lfloor \log_{n}n\rfloor\] is satisfied.