Choppin’ the Logs

For any integer $n \ge 2$ , let $S$ be the set $S \; = \; \big\{ (k,m) \in \mathbb{N}^2 \; \big| \; 1 \le k \le n \,,\,2 \le m \le n\,,\,k^m \le n \big\}$ Let us count the number of elements of $S$ in two different ways. Firstly $|S| \; = \; \sum_{m=2}^n \left|\big\{ k \in \mathbb{N} \,\big|\, 1 \le k \le n \,,\,k^m \le n \big\}\right| \; = \; \sum_{m=2}^n \left|\big\{ k \in \mathbb{N}\,\big|\, 1 \le k \le \sqrt[m]{n}\big\}\right| \; = \; \sum_{m=2}^n \left\lfloor \sqrt[m]{n}\right\rfloor$ On the other hand $\begin{aligned} |S| & = \; \sum_{k=1}^n \left|\big\{ m \in \mathbb{N}\,\big|\,2 \le m \le n\,,\,k^m \le n \big\}\right| \; = \; n-1 + \sum_{k=2}^n\left|\big\{ m \in \mathbb{N}\,\big|\, 2 \le m \le \log_k n\big\}\right| \\ & = \; n-1 + \sum_{k=2}^n \left( \left\lfloor \log_k n\right\rfloor - 1\right) \; = \; \sum_{k=2}^n \left\lfloor \log_k n\right\rfloor \end{aligned}$ Hence the identity in question is true for all integers $n \ge 2$ .