Prove that $\sum_{r=2}^{n} \left \lfloor n^{\frac{1}{r}} \right \rfloor = \sum_{r=2}^{n} \left \lfloor \log_{r}(n) \right \rfloor$.

Question

Prove that $$\sum_{r=2}^{n} \left \lfloor n^{\frac{1}{r}} \right \rfloor = \sum_{r=2}^{n} \left \lfloor \log_{r}(n) \right \rfloor\,.$$

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I have tried to use substitutions of $n=p^k$ in order to try and simplify, but it doesn't seem to lead anywhere, I'm not entirely sure how I can get around the floor functions.