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\begin{align*} \sum_{k=1}^{n} \frac{2^k-1}{k} &=\sum_{k=1}^{n} \frac{1}{k}\left(\sum_{j=1}^{k} \binom{k}{j}\right) \\ &=\sum_{j=1}^{n} \sum_{k=j}^{n} \binom{k}{j}\frac{1}{k} \\ &=\sum_{j=1}^{n} \frac{1}{j}\left(\sum_{k=j}^{n} \binom{k-1}{j-1}\right) \\ &=\sum_{j=1}^{n} \frac{1}{j} \binom{n}{j}. \end{align*}